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PhysicsCh 13 Oscillations

Oscillations: Class 11 Physics · Chapter 13

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 80 min
  • 24 practice questions
  • 24 MCQs

What you will learn in Oscillations

  • Distinguish periodic from oscillatory motion and relate period, frequency and angular frequency
  • Describe a simple harmonic motion by its amplitude, angular frequency, phase and phase constant, and decide whether a given function of time is simple harmonic, periodic or neither
  • Use the reference circle to find how the position, velocity and acceleration of an oscillating particle change with time, and compare their phases
  • State and apply the force law F = −kx, find the period of spring–mass systems, including vertical springs and springs in series or parallel
  • Calculate the kinetic, potential and total energy of a simple harmonic oscillator at any position
  • Derive the period of a simple pendulum for small swings and explain when the small-angle approximation holds
  • Solve board and entrance-level problems on SHM, including initial conditions and effective gravity

Original companion text aligned to NCERT Physics Class 11, Chapter 13, pp. 259–277. Not affiliated with or endorsed by NCERT.

13.1 Introduction

Some motions happen once and are over. A ball thrown to a fielder follows its arc and lands; a train runs from Pune to Mumbai and stops. Most of the motions in the earlier chapters of this course were of this one-off kind. Other motions come back to where they started, again and again: the Moon circling the Earth, a planet going round the Sun, the tip of a ceiling fan blade. Such motions are periodic.

This chapter looks at a special family of repeating motions in which a body goes back and forth on either side of a middle position. Drive over a speed breaker and the car keeps bobbing on its springs for a second or two. A diver leaves the end of a springboard and the board keeps quivering. Pluck a tanpura string and it blurs as it vibrates. Each of these is an oscillation (Fig. 13.1).

4/4Different objects, same pattern: to and fro about a middle position.

Fig. 13.1Three oscillators you may have seen today. Each moves to and fro on either side of a middle position.

Three panels, each with dashed outlines showing the two extreme positions and a double arrow for the to-and-fro motion. Step 1 highlights a car whose body bobs up and down on the springs above its wheels. Step 2 highlights a diving springboard fixed at one end whose free end flexes up and down. Step 3 highlights a string stretched between two pegs that bows up and down in the middle. Step 4 shows all three together: each moves back and forth on either side of a middle position.

Oscillations sit underneath a surprising amount of physics. Your eardrum moves to and fro whenever a sound reaches it. The quartz crystal that keeps time in a wristwatch vibrates tens of thousands of times a second. The voltage at a wall socket in India rises and falls through 50 complete cycles every second. And inside a solid that looks perfectly still, every atom is jiggling about its place in the crystal, more vigorously the hotter the solid is.

To describe any of these we need a small, exact vocabulary: period, frequency, displacement, amplitude and phase. The next section builds it, and the rest of the chapter uses it to study the simplest and most important oscillation of all, simple harmonic motion.

Quick check
1 of 2
Which of these motions does not repeat itself?
Show answer

Answer (A) a ball thrown once to a fielder

The throw happens once: the ball follows its arc and is caught. The Moon, the fan tip and the pendulum all come back to the same state again and again.

13.2 Periodic and oscillatory motions

Watch a car's turn indicator: on, off, on, off, with the same timing every time. Or follow the second hand of a wall clock, whose angle grows steadily for a minute and then starts again from twelve. If you plot what these do against time, the graph is made of one piece copied over and over (Fig. 13.2). That is the test for periodic motion: after a fixed interval, the motion repeats itself exactly.

3/4A complicated shape is still periodic if it repeats exactly.

Fig. 13.2Three periodic motions and one that is not. In each periodic graph a single piece, of duration T, is copied end to end.NCERT Fig. 13.1 · p. 260

Each step shows one graph on the same time axis. Step 1: an indicator light switches on and off, giving a row of identical rectangular pulses, with one repeat marked T. Step 2: the angle of a clock's second hand rises in a straight line and drops back to zero, a sawtooth, with one tooth marked T. Step 3: a heartbeat trace with a small bump, a tall spike and a rounded hump, repeated identically, one repeat marked T. Step 4: a swing left alone moves back and forth with a shrinking height, so no stretch of the graph ever repeats exactly.

Definition
Periodic motion

A motion that repeats itself exactly after equal intervals of time.

ExampleThe Earth's spin about its axis, which repeats every day.

Why do some repeating motions go back and forth? Drop a marble into a valley-shaped track (Fig. 13.3). Placed gently at the lowest point, it stays put: gravity and the push of the track cancel there, so this is its equilibrium position. Roll it a little way up either slope and the balance breaks; the leftover force always aims back down towards the bottom. That restoring force speeds the marble up on the way down, it overshoots, climbs the other side, is turned back again, and an oscillation is under way.

2/4Up the right slope, the net force points back towards the bottom.

Fig. 13.3A marble in a valley oscillates because every displacement brings a force back towards the bottom. On a hilltop the force points away, so there is no oscillation.

A curved track shaped like a valley. Step 1: a marble rests at the lowest point, labelled at rest with net force zero. Step 2: the marble sits up the right slope and an arrow along the track points back down towards the bottom, labelled pulled back. Step 3: the same on the left slope, with the arrow pointing right. Step 4: the valley is replaced by a hill; a marble just off the top has an arrow pointing further away from the top, labelled pushed away, no oscillation.

Definition
Oscillatory motion

A periodic to-and-fro motion of a body on either side of an equilibrium position, driven by a force that always points back towards that position.

ExampleA child's swing, or a weight bouncing on a spring.

Every oscillation is periodic, but the reverse is not true. The tip of a fan blade repeats its path every turn, yet it never reverses direction along a line, so it is periodic without being oscillatory. The same is true of a planet in its orbit.

Oscillation and vibration mean the same thing in physics. Habit picks the word: a tall building swaying in a storm, a few seconds per sway, is said to oscillate, while the prongs of a tuning fork, flicking back and forth hundreds of times a second, are said to vibrate.

Leave a swing alone and every arc is a little lower than the one before, until it hangs still. Friction at the hooks and the drag of the air drain its energy; this fading is called damping. A parent giving a well-timed push each cycle replaces the lost energy and keeps the swing going. For most of this chapter we ignore damping and study the cleanest case of all, simple harmonic motion, in which the restoring force grows in exact proportion to the distance from equilibrium. A real, slowly damped oscillator behaves like this only over stretches of time much shorter than the time it takes to die away.

Oscillators rarely live alone. Tie a row of beads along a stretched string and a jiggle given to the first bead is handed on to the next, and the next. A disturbance passed along a chain of linked oscillators in this way is a wave. Sound in air, the shaking of the ground after an earthquake and light are all waves, and they are the subject of the next chapter.

13.2.1 Period and frequency

The shortest time after which a periodic motion repeats is its period, written T. Its SI unit is the second. A motion that repeats after T also repeats after 2T, 3T and every whole-number multiple of T, but only the least such time is called the period.

Periods in nature span an enormous range. A tuning fork marked 512 Hz completes each vibration in about 2 milliseconds, while sunspots on the Sun wax and wane over a cycle of roughly 11 years and Saturn needs about 29 years to go once round the Sun. So it is often handier to say how many repeats happen each second. This number is the frequency, written ν (the Greek letter nu):

Frequency and period
ν=1T
(13.1)

Frequency is measured in s−1, repeats per second. This unit has its own name, the hertz (Hz):

The hertz
1 Hz=1 s−1
(13.2)

A frequency need not be a whole number. A pendulum that completes 7 swings in 10 s has ν=0.7 Hz; nothing requires a complete swing to finish exactly when a second ends.

Example 13.1NCERT Example 13.1 · p. 261

A fan and a crystal

(a) A ceiling fan turns steadily at 360 revolutions per minute. Find the frequency and period of its motion. (b) The quartz crystal in a wristwatch vibrates 32 768 times every second. Find its period.

Try it yourself first. Then reveal the solution one step at a time.

13.2.2 Displacement

Until now, displacement has meant a change of position. Oscillations need a wider meaning. Here the displacement is the amount by which whatever is repeating differs from its equilibrium value, at a given moment. Measuring from equilibrium is a choice, but almost always the most convenient one.

So the displacement of a block on a spring is its distance x from where the spring is relaxed, and that of a pendulum is most naturally the angle θ between the string and the vertical. It need not be a length or an angle at all. For the mains supply it is the voltage between the two pins of a socket, which swings positive and negative; for a sound reaching your ear it is how far the air pressure has risen above, or fallen below, its normal value (Fig. 13.4). Displacements carry a sign, and studying an oscillation means recording how the displacement changes with time.

5/5Different quantities, one idea: a displacement from equilibrium that repeats.

Fig. 13.4Displacement is any oscillating quantity measured from its equilibrium value: a position, an angle, a voltage or a pressure.NCERT Fig. 13.2 · p. 261

Four panels in a grid. Top left: a block on a spring next to a wall, displaced a distance x to the right of a dashed equilibrium line. Top right: a pendulum whose string makes an angle theta with a dashed vertical line. Bottom left: an AC source drawn as a circle with a wave sign, beside a small graph of its voltage rising and falling. Bottom right: a row of vertical lines for layers of air, bunched together in some places and spread out in others, labelled air pressure change. Steps 1 to 4 highlight one panel each; step 5 shows all four together.

Mathematically, a periodic displacement is a periodic function of time. The simplest ones are the sine and the cosine. Take

f(t)=Acos⁡ωt
(13.3a)

The cosine repeats whenever its argument grows by 2π. So f returns to the same value when ωt grows by 2π, that is, when t grows by

T=2πω
(13.3b)

and f(t+T)=f(t) for every t. The same holds for Asin⁡ωt. A mixture of the two with the same ω is periodic too, and it turns out to be a single shifted sine curve.

DerivationAdding a sine and a cosine of the same frequency

Start with the combination

f(t)=Asin⁡ωt+Bcos⁡ωt
(13.3c)

Choose two new constants D and ϕ so that A=Dcos⁡ϕ and B=Dsin⁡ϕ. Then f(t) becomes

Dsin⁡ωtcos⁡ϕ+Dcos⁡ωtsin⁡ϕ

which is D times the expansion of sin⁡(ωt+ϕ):

f=Dsin⁡(ωt+ϕ)
(13.3d)

Squaring and adding A=Dcos⁡ϕ and B=Dsin⁡ϕ, and dividing one by the other, gives the new constants:

D=A2+B2
tan⁡ϕ=BA

So the sum is a sine curve of the same period 2π/ω, with a bigger height D and shifted along the time axis by the angle ϕ (Fig. 13.5).

3/4The sum is a single sine curve: D=32+42=5 and tan⁡ϕ=4/3, so ϕ≈53∘.

Fig. 13.5Adding 3sin⁡ωt and 4cos⁡ωt gives 5sin⁡(ωt+53∘): one sine curve, same period. Adding many sines can build any periodic shape (Fourier).

Axes of value against omega t over two cycles, with grid lines at plus and minus 5. Step 1: a dashed blue sine curve of height 3. Step 2: a dotted orange cosine curve of height 4 is added. Step 3: the two components fade and a thick vermillion curve, their sum, appears with height 5, peaking a little earlier than the sine curve, shifted by 53 degrees. Step 4: the curves are replaced by a dashed square wave and a green curve made by adding four sines of frequencies omega, 3 omega, 5 omega and 7 omega, which already follows the square wave closely.

Example 13.2NCERT Example 13.2 · p. 262

Periodic or not?

Here ω is a positive constant. Which of these functions of time are periodic? Give the period of each periodic one. (i) 3sin⁡ωt+4cos⁡ωt (ii) cos⁡2ωt+sin⁡3ωt (iii) sin⁡ωt+sin⁡2 ωt (iv) e−ωtsin⁡ωt

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
Which of these motions is periodic but not oscillatory?
Show answer

Answer (D) the Moon going round the Earth

The Moon repeats its orbit, so its motion is periodic, but it does not move back and forth about an equilibrium point. The other three all reverse direction about a middle position, so they are oscillations as well as periodic.

13.3 Simple harmonic motion

Now for the star of the chapter. Put the equilibrium position at the origin of an x-axis. A body oscillating along this axis is in simple harmonic motion when a single cosine (or sine) curve describes its displacement at every instant:

Displacement in SHM
x(t)=Acos⁡(ωt+ϕ)
(13.4)
where
x(t)
displacement from equilibrium at time t
A
amplitude
ω
angular frequency
ωt+ϕ
phase, which changes with time
ϕ
phase constant, the phase at t=0
Definition
Simple harmonic motion (SHM)

Oscillation along a line in which the displacement from equilibrium is a sine or cosine function of time, x=Acos⁡(ωt+ϕ), with constant A, ω and ϕ.

The word simple is earned by that shape. Many motions repeat, but only those whose x–t graph is a pure sine curve qualify. A strobe photograph shows what such a motion looks like. Hang a weight on a spring, pull it up, let go, and photograph it with flashes at equal time intervals (Fig. 13.6). The images bunch up near the two ends, where the weight is slow, and spread out near the middle, where it is fast. Spread the flashes along a time axis and they trace a cosine curve.

3/3After one period T the weight is back at +A, and the dots trace a cosine curve.

Fig. 13.6Equal-time flashes bunch up at the ends of the swing and spread out in the middle. Plotted against time, they trace x=Acos⁡ωt.NCERT Figs. 13.4, 13.5 · p. 263

On the left a weight hangs from a spring, beside a column of tick marks for its positions. A dashed line marks equilibrium, with plus A above and minus A below. Step 1: the weight is held at plus A and one flash records it as a dot at time zero. Step 2: after half a cycle the weight is at minus A; seven flashes have been recorded, and the tick marks crowd together near the two ends and spread out in the middle; the dots on the time graph follow half a cosine curve. Step 3: a full cycle of thirteen dots, joined by a dashed cosine curve, ends back at plus A after one period T.

Each constant in Eq. (13.4) has a job. Start with A, the amplitude. A cosine never leaves the range from −1 to +1, so the body never strays further than A from the centre on either side: A is the farthest it gets, and by convention we keep it positive. Change A alone and the motion simply gets wider or narrower, with the turning points still reached at the same instants.

Next, the angle inside the cosine. Think of ωt+ϕ as a pointer that tells you how far through its cycle the body has got: whether it is at a turning point or crossing the centre, and in which direction. This angle is the phase, and it grows steadily with time. Where the pointer stood when the clock started, ϕ, is the phase constant (also called the phase angle). It is fixed by how the motion began: the starting displacement Acos⁡ϕ leaves two possible angles, and the starting direction picks one. Give two bodies the same A and ω but different ϕ and their graphs are the same curve slid along the time axis (Fig. 13.7).

3/4ϕ=−π/3 slides the whole curve later by T/6.

Fig. 13.7The three constants each change a different feature of the curve: A its height, ϕ its position along the time axis, ω how tightly it is packed.NCERT Figs. 13.7, 13.8 · p. 263

A dashed reference curve x equals A cos omega t runs over two periods, with marks at T and 2T. Step 1: only the reference curve. Step 2: a blue curve with 1.6 times the amplitude, crossing zero at the same times. Step 3: a green curve of the same height shifted later by one sixth of a period, phase constant minus pi by 3. Step 4: a purple curve with twice the angular frequency, completing two cycles in each period of the reference.

The third constant, ω, fixes the period.

DerivationAngular frequency and period

Measure time from a moment when the phase is zero, so that ϕ=0:

x(t)=Acos⁡ωt
(13.5)

After one period the displacement is the same again, x(t+T)=x(t):

Acos⁡ωt=Acos⁡ω(t+T)
(13.6)

The cosine first repeats when its argument has grown by 2π, so ω(t+T)=ωt+2π, which gives

ω=2πT=2πν
(13.7)

So ω, the angular frequency, is the rate at which the phase pointer turns, in radians per second. One full cycle is 2π radians, which is why ω is always 2π times the number of cycles per second. A faster pointer packs more cycles into the same time (step 4 of Fig. 13.7).

Example 13.3NCERT Example 13.3 · p. 264

SHM or merely periodic?

Decide whether each function of time represents (a) simple harmonic motion or (b) periodic motion that is not simple harmonic, and give each period: (1) 3cos⁡ωt+sin⁡ωt (2) sin⁡ωtcos⁡ωt (3) cos⁡ωt+cos⁡2ωt

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
A particle moves as x=0.05cos⁡(4πt+π/6) in SI units. What is its period?
Show answer

Answer (D) 0.5 s

ω=4π rad/s, so T=2π/ω=0.5 s. Taking 4π itself as the period, using π/ω, or quoting the frequency (2 Hz) as if it were a time gives the other options.

13.4 Simple harmonic motion and uniform circular motion

Uniform circular motion and SHM look very different, yet one hides inside the other. Look at a wheel turning steadily edge-on, with your eye in the plane of the wheel, and a point on the rim no longer seems to go round. It slides up and down along a straight line, slowing at the ends and rushing through the middle, exactly like an oscillator. The activity below lets you see this with a bicycle.

To turn this into algebra, three numbers describe the wheel: the radius A of the circle traced by the point P, its steady angular speed ω (anticlockwise, say), and the angle ϕ that the radius OP made with the positive x-axis when the clock started. The angle grows at a constant rate, so after a time t it has become ωt+ϕ (Fig. 13.8).

2/4OP has turned to ωt+ϕ, and P′ is at Acos⁡(ωt+ϕ).

Fig. 13.8As P goes round the reference circle, its foot P′ on a diameter moves in SHM. The projection on the y-axis is an SHM too, a quarter-cycle out of step.NCERT Fig. 13.10 · p. 265

A circle of radius A centred at O, with x and y axes. Step 1: at time zero the radius OP makes an angle phi with the x-axis; a dashed line drops from P to its foot P prime on the x-axis. Step 2: P has turned anticlockwise to a larger angle, omega t plus phi, and P prime has slid to the negative side of the axis. Step 3: P is below the axis and a double-headed arrow along the diameter shows that P prime only ever moves back and forth between minus A and plus A, with x equal to A cos of omega t plus phi. Step 4: a horizontal dashed line carries P's height across to a graph on the right, where the height traces a sine curve against time, y equal to A sin of omega t plus phi.

Drop a perpendicular from P to the x-axis and call its foot P′. The position of P′ is the x-component of OP:

x(t)=Acos⁡(ωt+ϕ)

This is exactly Eq. (13.4). Seen edge-on, so that only one coordinate counts, steady motion round a circle is simple harmonic motion, with the circle's radius as the amplitude, the angular speed as the angular frequency and the starting angle as the phase constant. The particle P is called the reference particle and its circle the reference circle.

Nothing makes the x-axis special. The shadow on the y-axis follows y(t)=Asin⁡(ωt+ϕ): the same swing and the same timing, but always a quarter of a cycle (π/2) away from the x-shadow. Turn this round and you get a useful fact: two SHMs along perpendicular lines, with equal amplitudes and a phase difference of π/2, combine into uniform circular motion.

Keep the geometry and the physics apart, though. The reference circle is a description of SHM, not its cause. A body in SHM along a line is pushed by a force along that line that grows with its distance from the centre. A body in uniform circular motion needs a centripetal force of constant size that keeps turning to point at the centre. The two forces are quite different.

Example 13.4NCERT Example 13.4 · p. 265

Shadows of two circular motions

Find the SHM of the x-projection of the particle in each case. (a) Radius 5 cm, period 2 s, anticlockwise; at t=0 the radius makes 60∘ with the positive x-axis, above the axis. (b) Radius 3 cm, period 8 s, clockwise; at t=0 the radius makes 135∘ with the positive x-axis (pointing up and to the left).

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
A reference particle goes round a circle of radius 4 cm once every 0.5 s. Its projection on a diameter moves in SHM with
Show answer

Answer (B) amplitude 4 cm and period 0.5 s

The projection swings from −4 cm to +4 cm, so the amplitude is the radius, not the diameter. It completes one swing for each turn, so the period is the same 0.5 s.

13.5 Velocity and acceleration in simple harmonic motion

The reference circle gives the velocity and acceleration of an oscillator almost for free. The reference particle P moves round a circle of radius A at angular speed ω, so its speed is

v=ωA
(13.8)

Its velocity points along the tangent to the circle. The oscillator P′ always sits directly below or above P on the x-axis, so the velocity of P′ is simply the x-component of the velocity of P (Fig. 13.9). When OP is at angle ωt+ϕ, the tangent makes the same angle with the vertical, and the x-component is −ωAsin⁡(ωt+ϕ). The minus sign says that while P is in the upper half of the circle, P′ moves towards negative x.

2/3Project the centripetal acceleration ω2A: a=−ω2Acos⁡(ωt+ϕ)=−ω2x.

Fig. 13.9The oscillator's velocity and acceleration are the x-components of the reference particle's velocity (tangent) and centripetal acceleration (towards O).NCERT Figs. 13.11, 13.12 · p. 266

A reference circle with P at 60 degrees and its foot P prime on the x-axis. Step 1: a green velocity arrow of length omega A along the tangent at P, and its horizontal component, pointing left, drawn at P prime. Step 2: a purple acceleration arrow of length omega squared A from P towards the centre, and its horizontal component, also pointing left, drawn at P prime. Step 3: two special positions. When P is at the end of the horizontal diameter its velocity is vertical, so P prime has zero velocity along x but the largest acceleration, pointing to the centre. When P is at the top of the circle, P prime is at the centre with the largest leftward velocity and zero acceleration.

DerivationVelocity and acceleration in SHM

Velocity. Projecting the tangent velocity of P gives

v(t)=−ωA×sin⁡(ωt+ϕ)
(13.9)

The same result follows without any geometry by differentiating Eq. (13.4), since velocity is the rate of change of displacement:

v(t)=ddtx(t)
(13.10)

Acceleration. The reference particle has a centripetal acceleration of size v2/A=ω2A, pointing from P to O. Its x-component is the acceleration of P′:

a(t)=−ω2A×cos⁡(ωt+ϕ)=−ω2x(t)
(13.11)

Differentiating the velocity once more gives the same thing:

a(t)=ddtv(t)
(13.12)

Read Eq. (13.11) slowly, because it is the heart of SHM: the acceleration is the displacement multiplied by the fixed negative number −ω2. Double the distance from the centre and the acceleration doubles; cross to the other side and it flips. So at every point of the swing the body is being urged back towards x=0, hardest at the turning points and not at all at the centre.

Eliminating the time between Eqs. (13.4) and (13.9) gives the speed at any position, a result used constantly in problems. Since sin⁡2+cos⁡2=1,

Speed at displacement x
v=± ωA2−x2

The speed is greatest, ωA, at the centre and zero at the two ends, just as the strobe photograph showed. With ϕ=0 the three quantities are x=Acos⁡ωt, v=−ωAsin⁡ωt and a=−ω2Acos⁡ωt. Stack their graphs on one time axis (Fig. 13.10) and slide a cursor along. All three share the period T; their peak values are A, ωA and ω2A. Each velocity peak arrives a quarter-cycle before the matching displacement peak (a phase lead of π/2), while the acceleration is always a mirror image of the displacement (a phase difference of π).

2/4x=0, v=−ωA, a=0: racing through the centre.

Fig. 13.10Displacement, velocity and acceleration in SHM share the period T. Velocity is a quarter-cycle ahead of displacement; acceleration is half a cycle away, always opposite to x.NCERT Fig. 13.13 · p. 267

Three graphs share one time axis running to T. Top: displacement, a cosine of height A. Middle: velocity, a negative sine of height omega A. Bottom: acceleration, a negative cosine of height omega squared A. A dashed vertical cursor marks the instant, with a dot on each curve. Step 1, t equal to 0: x is at plus A, v is zero and a is at minus omega squared A. Step 2, t equal to T by 4: x is zero, v is at its most negative and a is zero. Step 3, t equal to T by 2: x is at minus A, v is zero and a is at plus omega squared A. Step 4, t equal to 3T by 4: x is zero, v is at its most positive and a is zero.

InteractiveSHM explorer

A reference circle with a particle going round it, its projection on the horizontal diameter, and below it a block on a spring that moves with the projection. Graphs of displacement, velocity and acceleration against time share a moving cursor. Sliders set the amplitude, the period and the phase constant, and a mass slider sets the spring constant needed. Readouts show the phase, x, v, a, the kinetic and potential energy, and their constant total. Play starts the motion; a step button advances it by one twelfth of a period.

Fig. 13.11Set the amplitude, the period and the phase constant, then press Play. Watch the reference particle, the block on its spring and the three graphs move together.
Example 13.5NCERT Example 13.5 · p. 267

A bobbing float

A fishing float bobs up and down so that its displacement from its resting level is y=0.04cos⁡(5πt−π3), in metres, with t in seconds. At t=0.3 s find (a) the displacement, (b) the velocity and (c) the acceleration. (d) What are the greatest speed and the greatest acceleration?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
At which point of an SHM are the speed and the size of the acceleration both largest?
Show answer

Answer (D) at no point: when one is largest the other is zero

The speed is largest at the centre, where a=−ω2x=0; the acceleration is largest at the ends, where the particle is momentarily at rest. Choosing the mean position confuses large speed with large acceleration.

13.6 Force law for simple harmonic motion

Newton's second law turns the acceleration of Eq. (13.11) into a force. For a particle of mass m in SHM,

F=ma=−mω2x

Writing k for the positive constant mω2,

Force law for SHM
F(t)=−k x(t)
(13.13)
k=mω2
(13.14a)
ω=km
(13.14b)

so the period of a mass m held by a spring of spring constant k is

Period of a spring–mass oscillator
T=2πmk
LawForce law for SHM

A particle moves in simple harmonic motion exactly when the net force on it is proportional to its displacement from equilibrium and points the opposite way: F=−kx.

Because it always pulls the body back towards the mean position, this force is the restoring force. Notice that SHM can now be recognised in two ways. You may be given the motion, x=Acos⁡(ωt+ϕ), or you may be given the force, F=−kx. They carry the same information: two differentiations take you from the motion to the force, and two integrations (a skill for later) take you back, with A and ϕ appearing as the two constants of integration. In problems the force route is usually quicker: find the net force for a small displacement, and if it comes out as −(constant)×(displacement), the motion is SHM.

Only this force law gives SHM. A ball bouncing on a floor repeats its motion, but the force on it, steady gravity plus a sharp kick at each bounce, has nothing to do with −kx, so the motion is periodic without being simple harmonic. A body moving under a force exactly proportional to x is called a linear harmonic oscillator. Why is this one case so common? Near the bottom of almost any potential-energy valley, the force can be written as −kx plus smaller terms in x2, x3 and so on. For small swings those extra terms hardly matter, so atoms in a crystal, a floating boat and a swinging lamp all behave, to a good approximation, as linear oscillators. When the extra terms do matter, the oscillator is called non-linear.

Notice something remarkable in T=2πm/k: the amplitude does not appear. A spring oscillator takes the same time for a wide swing as for a narrow one, because a wider swing brings a proportionally larger restoring force. Planetary orbits behave quite differently. By Kepler's third law a bigger orbit takes longer, with T2 proportional to the cube of the orbit's size.

The block on a horizontal spring is the standard example, but the force law also covers springs that hang vertically. Gravity then does not spoil the SHM; it only moves the equilibrium point (Fig. 13.12).

3/3A further stretch y adds a spring force ky that weight does not balance: Fnet=−ky.

Fig. 13.12Hanging a mass on a spring only shifts the equilibrium down by d=mg/k. Measured from the new equilibrium, the net force is −ky: SHM with the same k.NCERT Figs. 13.14, 13.15 · p. 268

Three springs hang from the ceiling side by side. A dashed line marks the end of the unloaded spring, labelled natural length. Step 1: only the unloaded spring. Step 2: the middle spring carries a block and has stretched by d to a new equilibrium, marked by a second dashed line; a green arrow up labelled k d and a vermillion arrow down labelled m g balance, with k d equal to m g. Step 3: the right-hand block has been pulled a further distance y below the new equilibrium; the upward spring force k times d plus y is now larger than m g, and the net force is minus k y.

Example 13.6NCERT Example 13.6 · p. 267

A mass on a vertical spring

A 250 g mass hung from a light spring stretches it by 6.0 cm and rests there. It is then pulled down a little further and released. Show that it oscillates in SHM about its resting position, and find the period. Take g=9.8 m/s2.

Try it yourself first. Then reveal the solution one step at a time.

Two springs can share one mass. Side by side (in parallel), they stretch by the same amount and their forces add. End to end (in series), they carry the same force and their stretches add. Applying the force law to each arrangement gives the single spring constant that would do the same job:

kparallel=k1+k2
1kseries=1k1+1k2

For example, a 2.0 kg block held by springs of 300 N/m and 600 N/m in parallel oscillates with T=2π2.0/900≈0.30 s. With the same springs in series, k=200 N/m and T≈0.63 s. A block between two springs fixed to opposite walls is a parallel arrangement: a displacement stretches one spring and squashes the other, and both forces point back towards the middle.

Quick check
1 of 3
Which of these accelerations describe SHM? Pick all that apply.
Select all that apply
Show answer
Answer
  • (A) a=−36x
  • (D) a=−9(x−1)
SHM needs a proportional to the displacement from equilibrium and opposite to it. (A) is SHM with ω=6. (D) is SHM about x=1, with ω=3. (B) pushes the body away, and (C) is not proportional to x.

13.7 Energy in simple harmonic motion

An oscillator constantly trades one kind of energy for another. At the ends of the swing it stops for an instant, so it has no kinetic energy; at the centre it moves fastest. Meanwhile the spring is most stretched or squashed at the ends and relaxed at the centre. Each kind of energy rises from nothing to a peak and falls back again, while their sum never changes.

DerivationKinetic, potential and total energy

Kinetic energy. Put v from Eq. (13.9) into K=12mv2 and use mω2=k:

K=12kA2×sin⁡2(ωt+ϕ)
(13.15)

Squaring removes the sign of v, so K repeats twice in each oscillation: its period is T/2.

Potential energy. A potential energy can be defined only for a conservative force (Chapter 5). The restoring force −kx qualifies, and the work done against it in moving the body out to x from the centre, where we set U=0, is stored as

U=12kx2
(13.16)

With x from Eq. (13.4), this becomes

U=12kA2×cos⁡2(ωt+ϕ)
(13.17)

Total energy. Adding the two, the bracket sin⁡2+cos⁡2 equals 1:

E=K+U=12kA2
(13.18)

No t survives in Eq. (13.18): with no friction, nothing can drain or add energy, so the total is locked in from the moment of release. It grows as the amplitude squared, so doubling the swing stores four times the energy. Combining Eqs. (13.16) and (13.18) also gives the kinetic energy at any position, K=12k(A2−x2), which matches the speed formula of Section 13.5.

Plotted against x, the potential energy is a parabola, a bowl whose rim reaches the total-energy line exactly at x=±A (Fig. 13.13). At any position the gap between the bowl and the line is the kinetic energy. The particle can never go beyond ±A, because there U would exceed E and K would have to be negative, which is impossible. Plotted against time, K and U are two humps that take turns, each peaking twice per period.

2/4Here U=E/4 and K=3E/4: the bar splits into a quarter and three quarters.

Fig. 13.13Energy in SHM: the bowl U=12kx2 and the flat line E=12kA2. Kinetic energy fills the gap. In time, K and U trade places twice each period.NCERT Fig. 13.16 · p. 269

Step 1: a blue parabola U equal to half k x squared rises from zero at x equal to 0 to meet a horizontal vermillion line E equal to half k A squared at minus A and plus A. Step 2: at x equal to A by 2 a vertical bar from the axis up to E is split into a blue part U below the parabola, one quarter of the bar, and a vermillion part K above it, three quarters. Step 3: at plus A the whole bar is U; at the centre the whole bar is K. Step 4: a time graph over one period: a blue curve U starts at E and dips to zero at T by 4, a dashed vermillion curve K does the opposite, and the two always add to the flat line E; each peaks twice per period.

In both graphs nothing dips below zero. For K that is automatic, since it depends on v2. For U it is a consequence of our choice of zero at the mean position: only changes of U are physical, and putting the zero elsewhere would shift the bowl up or down without changing the motion.

Example 13.7NCERT Example 13.7 · p. 269

Energy of a sliding block

A 0.50 kg block on a smooth table is attached to a spring of spring constant 200 N/m. It is pulled 6.0 cm from equilibrium and released from rest. Find the potential, kinetic and total energy when the block is 2.0 cm from equilibrium, and the block's speed there. Where is the kinetic energy equal to the potential energy?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
The amplitude of a spring oscillator is doubled. Its total energy
Show answer

Answer (C) becomes four times as large

E=12kA2 grows as the square of the amplitude, so doubling A multiplies E by 4. 'Doubles' forgets the square.

13.8 The simple pendulum

Physicists strip a real pendulum down to an ideal, the simple pendulum: all the mass m sits in a bob small enough to treat as a point, the string of length L neither stretches nor weighs anything, and the support does not move. Released from one side, it swings back and forth in a single vertical plane. A nut on a thread comes surprisingly close.

Why are small swings simple harmonic? Pull the bob aside so that the string makes an angle θ with the vertical, and look at what acts on it: its weight mg, straight down, and the pull FT of the string, aimed at the support (Fig. 13.14). Which of these can bring the bob back? Break the weight into two parts, mgcos⁡θ along the line of the string and mgsin⁡θ along the arc of the swing.

3/4mgcos⁡θ along the string, mgsin⁡θ along the arc.

Fig. 13.14Along the string, FT−mgcos⁡θ supplies the centripetal force. Along the arc, mgsin⁡θ pulls the bob back towards the mean position and gives the restoring torque.NCERT Fig. 13.17 · p. 270

A bob hangs from a hatched support by a string of length L, pulled to the left of a dashed vertical line so that the string makes an angle theta with it. A dotted arc shows the bob's path. Step 1: just the displaced pendulum, with L and theta labelled. Step 2: a green arrow along the string towards the support, labelled F T for tension, and a vermillion arrow straight down labelled m g. Step 3: the weight is split into a dashed component m g cos theta pointing along the string away from the support and a purple component m g sin theta along the arc towards the mean position. Step 4: the arc from the bob to the mean position is highlighted and labelled s equal to L theta, and the purple component is labelled restoring torque about the support.

The string's pull and mgcos⁡θ both act along the string, straight through the support, so neither can turn the bob about the support. Their difference, FT−mgcos⁡θ, has a different job: it is the centripetal force mv2/L that keeps the bob on its circular arc. That leaves mgsin⁡θ, which always points along the arc towards the lowest point and does all the speeding up and slowing down. Since the bob turns about the support, torques about that point are the natural tool: the two forces along the string drop out at once.

DerivationPeriod of a simple pendulum

The tangential component mgsin⁡θ acts at a distance L from the support. Its torque tends to reduce θ, hence the minus sign:

τ=−L(mgsin⁡θ)
(13.19)

Newton's law for rotation about the support relates torque to angular acceleration α through the moment of inertia I about the support:

τ=Iα
(13.20)

Equating the two expressions for the torque,

Iα=−mgLsin⁡θ
(13.21)
α=−mgLIsin⁡θ
(13.22)

This is not yet SHM, because α is proportional to sin⁡θ, not to θ. But for an angle in radians,

sin⁡θ=θ−θ33!+θ55!−⋯
(13.23)

and when θ is small the second and later terms are tiny compared with the first. Keeping only θ,

α=−mgLI θ
(13.24)

Set this beside a=−ω2x (Eq. (13.11)): it is the same equation, with the angle θ as the displacement and mgL/I in the place of ω2. Small swings of a pendulum are therefore simple harmonic, with

ω=mgLIT=2πImgL
(13.25)

The string has no mass, so the moment of inertia is just that of the bob, I=mL2. The mass cancels:

Period of a simple pendulum
T=2πLg
(13.26)

How small is small? Compare θ in radians with sin⁡θ (Fig. 13.15 and Table 13.2). Up to about 15∘ the two differ by little more than one per cent, and at 25∘ by about three per cent; by 60∘ the gap is large.

3/3Below about 20∘ the two are nearly equal; at 60∘ they differ by 0.18.

Fig. 13.15sin⁡θ and θ (in radians) are almost equal for small angles, which is what makes a pendulum's small swings simple harmonic.NCERT Table 13.1 · p. 271

Axes of value against angle from 0 to 90 degrees, with grid lines at 20, 40, 60 and 80 degrees and at 0.5, 1.0 and 1.5. Step 1: a dashed blue straight line for theta in radians, reaching about 1.57 at 90 degrees. Step 2: a vermillion curve for sine theta, which hugs the line at first and then bends below it, reaching 1 at 90 degrees. Step 3: the region below 20 degrees is shaded and labelled nearly equal, and a double arrow at 60 degrees marks the gap of 0.18 between the line and the curve.

Table 13.2How closely sin⁡θ follows θ (in radians)NCERT Table 13.1 · p. 271
θ (degrees)θ (radians)sin⁡θdifference
40.06980.06980.08 %
80.13960.13920.33 %
120.20940.20790.73 %
160.27930.27561.3 %
250.43630.42263.2 %
400.69810.64288.6 %

A period that does not depend on the amplitude is exactly what a clock needs. For a real pendulum this is only approximately true, and the lab below lets you test it. A swing of 20∘ on either side takes about 0.8 per cent longer than Eq. (13.26) predicts, and a 60∘ swing about 7 per cent longer. Galileo's lamp, swinging through a few degrees, was well inside the safe range.

InteractivePendulum lab

A pendulum swings from a support, solved exactly from the full equation with sin theta, beside a faint grey pendulum that follows the small-angle SHM formula. Sliders set the string length and the release angle, and buttons choose the Earth, the Moon or Mars. Play starts both pendulums; the panel shows the measured period, the small-angle prediction two pi root L over g, their percentage difference and the number of swings completed.

Fig. 13.16Choose the length, the release angle and the planet, then press Play. Compare the measured period with 2πL/g, and watch the grey small-angle pendulum drift ahead at large angles.

Eq. (13.25) is more general than the simple pendulum. Any rigid body swinging about a fixed pivot, such as a ruler hung from a nail through one end, obeys it for small swings, with I its moment of inertia about the pivot and L the distance from the pivot to its centre of mass. Such a body is called a physical pendulum.

Example 13.8NCERT Example 13.8 · p. 271

A half-second pendulum

A clockmaker wants a pendulum with a period of exactly 1.00 s, so that it ticks every half-second. (a) How long must it be? Take g=9.80 m/s2. (b) By what factor must the length change to double the period, and how long is that pendulum?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
A simple pendulum is moved from the Earth to the Moon, where g is about one-sixth as large. Its period
Show answer

Answer (B) becomes about 2.4 times longer

T∝1/g, so the period grows by 6≈2.4. 'Six times' forgets the square root; 'shorter' gets the dependence upside down.
Recap

Oscillations: key points and summary

  1. A motion that repeats itself exactly after equal intervals is periodic; an oscillation is a periodic to-and-fro motion about an equilibrium position (Section 13.2).
  2. The period T is the least time for one repeat; the frequency is ν=1/T, measured in hertz (1 Hz=1 s−1) (Section 13.2.1).
  3. Displacement means any oscillating quantity measured from its equilibrium value; Asin⁡ωt+Bcos⁡ωt is a single sinusoid Dsin⁡(ωt+ϕ), and any periodic function is a sum of sinusoids (Fourier) (Section 13.2.2).
  4. In SHM, x(t)=Acos⁡(ωt+ϕ): amplitude A, phase ωt+ϕ, phase constant ϕ, angular frequency ω=2π/T=2πν (Section 13.3).
  5. A point moving steadily round a circle, viewed along one diameter, moves in SHM: the reference circle links the two motions (Section 13.4).
  6. v=−ωAsin⁡(ωt+ϕ) and a=−ω2x; the maxima are ωA and ω2A, and v=±ωA2−x2 (Section 13.5).
  7. Velocity leads displacement by π/2; acceleration is opposite to displacement (phase difference π) (Section 13.5).
  8. SHM happens exactly when F=−kx: then ω=k/m and T=2πm/k, independent of amplitude (Section 13.6).
  9. Gravity only shifts the equilibrium of a vertical spring; springs combine as k1+k2 in parallel and k1k2/(k1+k2) in series (Section 13.6).
  10. K=12k(A2−x2) and U=12kx2 each have period T/2; their total E=12kA2 stays constant (Section 13.7).
  11. For small angles a simple pendulum moves in SHM with T=2πL/g, independent of the bob's mass (Section 13.8).
  12. The approximation sin⁡θ≈θ is good to about 1 % up to 15∘; larger swings take longer than the formula predicts (Section 13.8).
Practice · 24 questions

Practice questions with answers (NCERT “Exercises”)

Questions 1–18 practise the same skills as the NCERT exercises, in new settings. Questions 19–24 are extra objective items in the style of entrance examinations.

  1. Quick answerEasyNCERT Exercise 13.1

    Which of these motions are periodic? Pick all that apply.

    Select all that apply
    Show answer
    Answer
    • (A) the tip of a ceiling fan blade turning at a steady speed
    • (C) a child on a swing given one push, ignoring friction
    • (E) the hour hand of a wall clock
    The fan tip, the frictionless swing and the hour hand all repeat their motion at fixed intervals. The stone's flight happens once and stops, and the rocket never returns, so neither is periodic.
  2. Quick answerMediumNCERT Exercise 13.2

    Which of these motions are (nearly) simple harmonic? The others are periodic but not simple harmonic. Pick all the (nearly) simple harmonic ones.

    Select all that apply
    Show answer
    Answer
    • (B) a test tube weighted with sand, floating upright in water, pushed down a little and released
    • (D) a playground swing making small swings
    • (E) a trolley held between two springs on a smooth track, pulled aside and released
    The floating tube (extra buoyancy proportional to the extra depth), the swing at small angles and the trolley between springs all have restoring forces proportional to displacement. The merry-go-round repeats but does not move to and fro along a line, and the bouncing ball moves under constant gravity with sudden reversals, so both are periodic but not simple harmonic.
  3. Quick answerEasyNCERT Exercise 13.3

    Each graph in the figure shows the displacement x of a particle moving along a line against time t. Decide whether each motion is periodic, and give the period of each periodic one.

    Fig. 13.17Four displacement–time graphs for Exercise 3.

    Four small graphs of displacement against time in seconds. Graph a: a zigzag of straight rising and falling lines, with peaks at 2 and 6 seconds and troughs at 0, 4 and 8 seconds. Graph b: a curve that rises ever more steeply and never comes back down. Graph c: a pattern of one tall narrow bump followed by one low wide bump, the same pair repeated every 3 seconds. Graph d: an oscillation whose peaks get smaller and smaller as time goes on.

    1. Graph (a)

      Show answer

      Answer (B) periodic, T=4 s

      The zigzag repeats from one trough to the next: troughs at 0, 4 and 8 s, so T=4 s. The 2 s option is only from a trough to the next peak.
    2. Graph (b)

      Show answer

      Answer (C) not periodic

      The displacement keeps growing and never returns to an earlier value, so nothing repeats.
    3. Graph (c)

      Show answer

      Answer (B) periodic, T=3 s

      The tall bump and the low bump together form the repeating unit, which lasts 3 s. The two bumps differ, so half of that is not a repeat.
    4. Graph (d)

      Show answer

      Answer (C) not periodic

      The swings shrink each time, so no stretch of the graph is ever repeated exactly. It is a damped oscillation.
  4. Quick answerMediumNCERT Exercise 13.4

    For each function of time, decide whether it represents simple harmonic motion, periodic motion that is not simple harmonic, or non-periodic motion. (ω is a positive constant.)

    1. 2sin⁡ωt+2cos⁡ωt

      Show answer

      Answer (A) simple harmonic

      It equals 22sin⁡(ωt+π/4): SHM with period 2π/ω.
    2. 1−2sin⁡2ωt

      Show answer

      Answer (A) simple harmonic

      The double-angle formula turns it into cos⁡2ωt, a single sinusoid: SHM with period 2π/2ω=π/ω.
    3. cos⁡ωtcos⁡2ωt

      Show answer

      Answer (B) periodic but not simple harmonic

      The product-to-sum formula gives 12(cos⁡ωt+cos⁡3ωt), two different frequencies. It repeats with period 2π/ω but is not SHM.
    4. ωtcos⁡ωt

      Show answer

      Answer (C) not periodic

      The factor ωt makes each swing larger than the last, so the function never repeats.
  5. Short answerMediumNCERT Exercise 13.5

    A particle moves in SHM along a straight line between points P and Q, 12 cm apart. The direction from P to Q is positive. Give the signs (positive, negative or zero) of the particle's velocity, acceleration and the force on it when it is:

    1. at Q

      Show solution

      Answerv=0; a and F negative.

      At an end the particle is momentarily at rest, so v=0. It is at x=+6 cm, and a=−ω2x points back towards O, the negative direction. The force has the same sign as a.

    2. at the midpoint, moving towards Q

      Show solution

      Answerv positive; a=0 and F=0.

      At O, x=0, so a=F=0. Moving towards Q means moving in the positive direction.

    3. 3 cm from P, moving towards P

      Show solution

      Answerv negative; a and F positive.

      Here x=−3 cm (P is at −6 cm). Moving towards P is the negative direction. With x<0, a=−ω2x>0.

    4. 2 cm from Q, moving towards P

      Show solution

      Answerv negative; a and F negative.

      Here x=+4 cm, so a and F point towards O: negative. Moving towards P makes v negative too. Velocity and acceleration need not have opposite signs; they share a sign whenever the particle is heading back towards the centre.

  6. Quick answerEasyNCERT Exercise 13.6

    For which of these rules linking a particle's acceleration a to its position x does the particle move in SHM? Pick all that apply.

    Select all that apply
    Show answer
    Answer
    • (A) a=−25x
    • (B) a=4−9x
    (A) is SHM with ω=5 rad/s. (B) can be written a=−9(x−49): SHM about x=49 with ω=3 rad/s. In (C) the acceleration always points to the origin but has the same size everywhere, so it is periodic but not SHM. In (D) the acceleration points away from equilibrium, so there is no oscillation at all.
  7. NumericalHardNCERT Exercise 13.7

    A particle moves in SHM with angular frequency π rad/s. At t=0 its displacement is −3 cm and its velocity is +π cm/s.

    1. Writing x=Acos⁡(ωt+ϕ), find A and ϕ.

      Show solution

      AnswerA=2 cm, ϕ=−5π/6 (equivalently 7π/6).

      Acos⁡ϕ=−3−Aωsin⁡ϕ=π

      With ω=π the second gives Asin⁡ϕ=−1. Squaring and adding, A2=3+1, so A=2 cm. Then cos⁡ϕ=−3/2 and sin⁡ϕ=−12, which puts ϕ in the third quadrant: ϕ=−5π/6.

    2. Writing instead x=Bsin⁡(ωt+α), find B and α.

      Show solution

      AnswerB=2 cm, α=−π/3.

      Bsin⁡α=−3Bωcos⁡α=π

      So Bcos⁡α=1 and again B=2 cm. Then sin⁡α=−3/2 and cos⁡α=12, so α=−π/3. As expected, α=ϕ+π/2: a sine is a cosine shifted by a quarter-cycle.

  8. NumericalMediumNCERT Exercise 13.8

    A 1.0 kg packet of rice hung from a spring stretches it by 4.0 cm. The rice is replaced by a watermelon, which is pulled down slightly and released; it bounces with a period of 0.70 s. What is the weight of the watermelon, in newtons? Take g=9.8 m/s2.

    N
    Show solution

    From the static stretch,

    k=(1.0)(9.8)0.040=245 N/m

    Rearranging T=2πm/k:

    m=kT24π2=245(0.49)39.5
    m=3.04 kg

    The weight is mg=29.8 N. A spring can work as a balance through its period as well as through its stretch.

  9. NumericalEasyNCERT Exercise 13.9

    A 2.0 kg block on a smooth horizontal table is fixed to a wall by a spring of spring constant 800 N/m. It is pulled 3.0 cm from its equilibrium position and released.

    1. What is the frequency of the oscillations, in hertz?

      Hz
      Show answer

      Answer 3.18 Hz

      ω=800/2.0=20 rad/s, so ν=20/2π=3.18 Hz.
    2. What is the greatest acceleration of the block, in m/s2?

      m/s²
      Show answer

      Answer 12 m/s²

      amax⁡=ω2A=400(0.030)=12 m/s2, at the ends of the swing.
    3. What is the greatest speed of the block, in m/s?

      m/s
      Show answer

      Answer 0.6 m/s

      vmax⁡=ωA=20(0.030)=0.60 m/s, at the equilibrium position.
  10. Short answerMediumNCERT Exercise 13.10

    For the block of Question 9, take x=0 where the spring is relaxed, with x positive when the spring is stretched. Write x(t), in cm, if at the moment the stopwatch is started (t=0) the block is:

    1. at the point of greatest compression

      Show solution

      Answerx=3cos⁡(20t+π) cm=−3cos⁡20t cm

      x(0)=−3 needs cos⁡ϕ=−1, so ϕ=π.

    2. passing through x=0 while moving in the positive direction

      Show solution

      Answerx=3cos⁡(20t−π/2) cm=3sin⁡20t cm

      x(0)=0 needs cos⁡ϕ=0, and v(0)>0 needs sin⁡ϕ<0, so ϕ=−π/2.

    3. at x=+1.5 cm, moving away from equilibrium

      Show solution

      Answerx=3cos⁡(20t−π/3) cm

      cos⁡ϕ=12 gives ϕ=±π/3. Moving away from equilibrium at positive x means v(0)>0, so sin⁡ϕ<0 and ϕ=−π/3.

    4. In what ways do the three functions differ?

      Show solution

      AnswerOnly in the phase constant.

      All three have amplitude 3 cm and angular frequency 20 rad/s, which are fixed by the spring, the mass and the release. Only ϕ, which depends on when the stopwatch starts, is different.

  11. Short answerMediumNCERT Exercise 13.11

    A particle moves round a circle at a steady rate. In each case write the SHM of the x-projection of its position.

    1. Radius 4 cm, period 3 s, anticlockwise; at t=0 the particle is on the positive y-axis.

      Show solution

      Answerx=4cos⁡(2πt/3+π/2)=−4sin⁡(2πt/3) cm

      ω=2π/3 rad/s and the start angle is π/2; anticlockwise, the angle at time t is 2πt/3+π/2. So x=4cos⁡(2πt/3+π/2) cm: amplitude 4 cm, period 3 s, phase constant π/2. The projection starts at the centre and moves towards negative x.

    2. Radius 1.5 m, period 6 s, clockwise; at t=0 the radius points 30∘ below the positive x-axis.

      Show solution

      Answerx=1.5cos⁡(πt/3+π/6) m

      ω=π/3 rad/s and the start angle is −π/6. Clockwise, the angle at time t is −π/6−πt/3. Since cosine is even, x=1.5cos⁡(πt/3+π/6) m: amplitude 1.5 m, period 6 s, phase constant π/6.

  12. Short answerHardNCERT Exercise 13.12

    For each SHM below (x in cm, t in s), describe the reference circle: its radius, the angular speed of the reference particle and the particle's starting angle, measured anticlockwise from the positive x-axis. Take the rotation to be anticlockwise in every case.

    1. x=3sin⁡2t

      Show solution

      AnswerRadius 3 cm, 2 rad/s, start angle −π/2 (bottom of the circle).

      sin⁡u=cos⁡(u−π/2), so 3sin⁡2t=3cos⁡(2t−π/2). Turning anticlockwise from the bottom, the projection first moves towards positive x, as 3sin⁡2t requires.

    2. x=−2cos⁡(t/2)

      Show solution

      AnswerRadius 2 cm, 0.5 rad/s, start angle π (left end of the diameter).

      −cos⁡u=cos⁡(u+π), so x=2cos⁡(t/2+π).

    3. x=sin⁡πt+cos⁡πt

      Show solution

      AnswerRadius 2 cm, π rad/s, start angle −π/4 (45∘ below the x-axis).

      Combine as in Eq. (13.3d): sin⁡πt+cos⁡πt=2cos⁡(πt−π/4).

    4. x=4sin⁡(2πt+5π/6)

      Show solution

      AnswerRadius 4 cm, 2π rad/s, start angle π/3 (60∘).

      4sin⁡(2πt+5π/6)=4cos⁡(2πt+5π/6−π/2)=4cos⁡(2πt+π/3).

  13. NumericalHardNCERT Exercise 13.13

    A light spring has spring constant 150 N/m. In set-up (i) one end is fixed to a wall and a 0.30 kg trolley is attached to the other; a 6.0 N pull on the trolley stretches the spring. In set-up (ii) the same spring joins a 0.30 kg trolley to a 0.60 kg trolley on a smooth track, and each trolley is pulled outwards with 6.0 N. The trolleys are then released.

    1. What is the extension of the spring in each set-up?

      Show solution

      Answer4.0 cm in both.

      In both set-ups the tension in the spring is 6.0 N: in (ii) the spring is pulled by 6.0 N at each end, just as the wall and the trolley pull it in (i). So the extension is F/k=6.0/150=0.040 m in both.

    2. How long does one oscillation take in each set-up?

      Show solution

      Answer(i) 0.28 s; (ii) 0.23 s.

      (i) T=2π0.30/150=0.28 s.

      (ii) No external horizontal force acts, so the centre of mass stays still. It divides the spring in the inverse ratio of the masses: the 0.30 kg trolley is attached to 2/3 of the spring. A piece of spring 2/3 as long is 3/2 as stiff, so that trolley oscillates on an effective spring of 225 N/m:

      T=2π0.30225=0.23 s

      The other trolley, on 1/3 of the spring (450 N/m), gives the same period, 2π0.60/450, as it must. This is 2πμ/k with μ=m1m2/(m1+m2)=0.20 kg.

  14. NumericalEasyNCERT Exercise 13.14

    The needle of a sewing machine moves up and down with a stroke (twice the amplitude) of 2.8 cm, making 900 stitches per minute. Treating its motion as SHM, what is its greatest speed, in m/s?

    m/s
    Show solution
    ω=2π(15)=94.2 rad/s
    vmax⁡=ωA=94.2(0.014)
    vmax⁡=1.32 m/s
  15. NumericalEasyNCERT Exercise 13.15

    A pendulum has a period of 2.4 s on the Earth (g=9.8 m/s2). What would its period be on Mars, where g=3.7 m/s2? Answer in seconds.

    s
    Show solution

    With L fixed, T∝1/g:

    TMars=TEgEgM
    TMars=2.49.83.7=3.9 s
  16. NumericalHardNCERT Exercise 13.16

    A pendulum 0.80 m long hangs from the roof of a bus. Take g=9.8 m/s2. Find its period of small oscillations when:

    1. the bus accelerates along a straight, level road at a steady 2.5 m/s2 (the bob swings in the direction of travel)

      Show solution

      Answer1.77 s (compared with 1.80 s at rest).

      In the bus's frame the bob feels gravity plus a backward pseudo-force ma, at right angles to each other:

      geff=9.82+2.52=10.1 m/s2
      T=2π0.8010.1=1.77 s

      The string hangs tilted backwards by tan⁡−1(2.5/9.8)≈14∘, and the bob oscillates about that tilted line.

    2. the pendulum is instead taken into a lift that accelerates upwards at 2.0 m/s2

      Show solution

      Answer1.64 s

      Now the pseudo-force points straight down, adding to gravity: geff=9.8+2.0=11.8 m/s2, so T=2π0.80/11.8=1.64 s.

  17. ProofHardNCERT Exercise 13.17

    A uniform test tube, weighted at the bottom so that it floats upright in water, floats with a length d submerged. It is pushed down slightly and released. Ignoring the drag of the water and the motion of the water itself:

    1. Show that the tube moves in SHM with period T=2πd/g.

      Show solution

      AnswerNet force −ρlAgy, so T=2πm/ρlAg=2πd/g.

      Let the tube have mass m and cross-section A, and the liquid density ρl. Floating at rest, ρlAdg=mg. Pushed down a further y, the buoyancy becomes ρlA(d+y)g, so the net upward force is

      F=−ρlAg y

      (taking y positive downwards). This has the form F=−ky with k=ρlAg, so the motion is SHM with

      T=2πmρlAg=2πdg

      using m=ρlAd. For a solid cylinder of height h and density ρ, d=hρ/ρl, so T=2πhρ/ρlg.

    2. Find the period, in seconds, when 8.0 cm of the tube is under water.

      s
      Show answer

      Answer 0.568 s

      T=2π0.080/9.8=0.57 s.
  18. ProofHardNCERT Exercise 13.18

    Water stands in a U-shaped glass tube of uniform bore. The total length of the water column, measured along the tube, is L. You blow into one arm to push its level down a little, then stop.

    1. Show that the water column moves in SHM, and find its period.

      Show solution

      AnswerF=−2ρAg y, so T=2πL/2g.

      Let the bore have cross-section A and the water density ρ; the moving mass is m=ρAL. When one level is y below its rest position, the other is y above, so a column of height 2y is unbalanced. Its weight pushes the whole column back:

      F=−(2y)Aρg=−2ρAg y

      This is a restoring force proportional to y, so the motion is SHM with k=2ρAg:

      T=2πρAL2ρAg=2πL2g

      The density cancels, so mercury and water columns of the same length oscillate equally fast.

    2. Find the period, in seconds, for L=60 cm.

      s
      Show answer

      Answer 1.1 s

      T=2π0.60/19.6=1.10 s.
  19. Quick answerMedium

    Assertion (A): The period of a simple pendulum does not depend on the mass of its bob. Reason (R): Both the restoring force on the bob and its inertia are proportional to its mass.

    Show answer

    Answer (A) Both A and R are true, and R explains A

    The restoring torque is mgLθ and the moment of inertia is mL2; the mass cancels in their ratio, which is exactly why the period is independent of mass.
  20. Quick answerMedium

    Assertion (A): In SHM the speed is greatest at the mean position. Reason (R): In SHM the acceleration is greatest at the mean position.

    Show answer

    Answer (C) A is true but R is false

    A is true: v=ωA2−x2 is largest at x=0. R is false: a=−ω2x is zero at the mean position and largest at the extremes.
  21. Quick answerMedium

    A particle in SHM has amplitude 5 cm. Its speed is 12 cm/s when it is 3 cm from the centre. What is its period, in seconds?

    s
    Show answer

    Answer 2.094 s

    12=ω25−9=4ω, so ω=3 rad/s and T=2π/3=2.09 s.
  22. Quick answerHard

    A particle in SHM with period 2.4 s starts from the mean position. How long, in seconds, does it take to first reach half its amplitude?

    s
    Show answer

    Answer 0.2 s

    sin⁡ωt=12 first at ωt=π/6, so t=T/12=0.20 s. It is not T/8: the particle is fastest near the centre, so the first half of the distance takes less than half of the quarter-period T/4.
  23. Quick answerEasy

    Match each quantity in an SHM x=Acos⁡ωt with how it behaves.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • velocity : a quarter-cycle ahead of x
    • acceleration : always opposite in sign to x
    • kinetic energy : largest at x=0, period T/2
    • total energy : constant, 12kA2
    v=−ωAsin⁡ωt=ωAcos⁡(ωt+π/2); a=−ω2x; K=12k(A2−x2) peaks at the centre twice per period; K+U=12kA2.
  24. Quick answerMedium

    A block on a spring oscillates with period T. The spring is cut into two equal halves and the same block is hung from one half. The new period is

    Show answer

    Answer (C) T/2

    Half the spring has twice the spring constant, and T∝1/k, so the period becomes T/2. T/2 forgets the square root; 2 T gets the stiffness change backwards.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercise 13.1Practice Q1
NCERT Exercise 13.2Practice Q2
NCERT Exercise 13.3Practice Q3
NCERT Exercise 13.4Practice Q4
NCERT Exercise 13.5Practice Q5
NCERT Exercise 13.6Practice Q6
NCERT Exercise 13.7Practice Q7
NCERT Exercise 13.8Practice Q8
NCERT Exercise 13.9Practice Q9
NCERT Exercise 13.10Practice Q10
NCERT Exercise 13.11Practice Q11
NCERT Exercise 13.12Practice Q12
NCERT Exercise 13.13Practice Q13
NCERT Exercise 13.14Practice Q14
NCERT Exercise 13.15Practice Q15
NCERT Exercise 13.16Practice Q16
NCERT Exercise 13.17Practice Q17
NCERT Exercise 13.18Practice Q18

Quick check: MCQs on Oscillations

  1. 13.1 Introduction2 questions
  2. 13.2.2 Displacement3 questions
  3. 13.3 Simple harmonic motion3 questions
  4. 13.4 Simple harmonic motion and uniform circular motion3 questions
  5. 13.5 Velocity and acceleration in simple harmonic motion3 questions
  6. 13.6 Force law for simple harmonic motion3 questions
  7. 13.7 Energy in simple harmonic motion3 questions
  8. 13.8 The simple pendulum4 questions
Glossary · 14

Key terms (English–Hindi)

amplitudeआयामaayaam
The largest magnitude of the displacement from equilibrium in an oscillation, A.First seen in the chapter →
angular frequencyकोणीय आवृत्तिkoniya aavritti
ω=2π/T=2πν, in radians per second.First seen in the chapter →
frequencyआवृत्तिaavritti
The number of repeats per second, ν=1/T, measured in hertz (Hz).First seen in the chapter →
linear harmonic oscillatorरैखिक आवर्ती दोलित्रraikhik aavartee dolitra
A body acted on by a restoring force exactly proportional to its displacement.First seen in the chapter →
oscillatory motionदोलन गतिdolan gati
Periodic to-and-fro motion about an equilibrium position, caused by a force that points back towards it.First seen in the chapter →
periodआवर्तकालaavartkaal
The least time after which a periodic motion repeats, T; SI unit second.First seen in the chapter →
periodic motionआवर्ती गतिaavartee gati
Motion that repeats itself exactly after equal intervals of time.First seen in the chapter →
phaseकलाkala
The angle ωt+ϕ in x=Acos⁡(ωt+ϕ), which fixes the position and direction of motion at time t.First seen in the chapter →
phase constantकला स्थिरांकkala sthirank
The value ϕ of the phase at t=0.First seen in the chapter →
reference circleनिर्देश वृत्तnirdesh vritt
A circle round which a particle moves uniformly; its projection on a diameter performs SHM.First seen in the chapter →
restoring forceप्रत्यानयन बलpratyanayan bal
A force that always points back towards the equilibrium position; in SHM it is −kx.First seen in the chapter →
simple harmonic motionसरल आवर्त गतिsaral aavart gati
Oscillation in which the displacement is a sinusoidal function of time, produced by a force F=−kx.
simple pendulumसरल लोलकsaral lolak
A point-like bob on a light, inextensible string; for small swings T=2πL/g.First seen in the chapter →
spring constantस्प्रिंग नियतांकspring niyatank
The force per unit extension of a spring, k, in N/m; in SHM k=mω2.First seen in the chapter →