NCERT Solutions Class 11 Physics Chapter 13 — Oscillations

NCERT Class 11 Physics — Oscillations, Chapter 13 Exercises. All 18 questions solved.

The exercises at the end of Chapter 13 fall into three groups. Questions 13.1 to 13.6 ask you to recognise motion: which is periodic, which is simple harmonic, and which neither. Questions 13.7 to 13.12 work with the equation of simple harmonic motion (SHM) itself — its amplitude, phase and period, and the reference circle that generates it. Questions 13.13 to 13.18 are physical oscillators: springs, pendulums, a floating cork and a column of mercury. Everything runs on three results:

• The SHM equation — $x(t) = A\cos(\omega t + \phi)$ with $\omega = 2\pi/T = 2\pi\nu$, so that $v = -\omega A\sin(\omega t + \phi)$ and $a = -\omega^2 x$. The greatest speed is $\omega A$ and the greatest acceleration $\omega^2 A$.
• A linear restoring force — $F = -kx$ produces SHM with $\omega = \sqrt{k/m}$, that is $T = 2\pi\sqrt{m/k}$.
• The simple pendulum — for small swings, $T = 2\pi\sqrt{L/g}$.

Key insight. A motion is simple harmonic exactly when its acceleration is proportional to the displacement and directed opposite to it, $a = -\omega^2 x$. That single test settles every identification question (13.2, 13.4, 13.6), and every oscillator question (13.8, 13.13, 13.17, 13.18) comes down to writing the restoring force in the form $F = -kx$: once $k$ is found, $T = 2\pi\sqrt{m/k}$ follows at once.

Question 13.1

Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow.

Solution. A motion is periodic if it repeats itself at regular intervals of time. (a) The swimmer makes a single return trip and stops; nothing is repeated, so it is not periodic, even though the swimmer ends where the trip began. (b) The magnet swings back and forth about the N-S direction, each swing taking the same time, so it is periodic. (c) The molecule turns about its centre of mass at a steady rate and returns to the same orientation after each revolution, so it is periodic. (d) The arrow flies off and never returns, so it is not periodic.

Question 13.2

Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position.

Solution. SHM is a to-and-fro motion about an equilibrium position under a restoring force proportional to the displacement.

(a) The earth’s rotation repeats every day, so it is periodic, but it is not a to-and-fro motion about an equilibrium position at all, and there is no restoring force. (b) If the mercury on one side is pushed down by $y$, the other side rises by $y$, and the extra weight of a column of height $2y$ pushes it back — a restoring force proportional to $y$. This is SHM, shown in detail in Question 13.18. (c) Near the bottom of a curved bowl, the component of gravity along the surface is proportional to the displacement from the lowest point, exactly as for a pendulum bob, so for a small release the ball executes SHM. (d) A polyatomic molecule has several natural frequencies of vibration. Its general motion is a superposition of SHMs of these different frequencies, which repeats but is not itself a single SHM.

Question 13.3

Fig. 13.18 depicts four $x$–$t$ plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

The caption under this figure in the book reads “Fig. 18.18”, a misprint for Fig. 13.18.

Solution. A motion is periodic if the whole pattern of positions repeats at regular intervals, not merely if the particle comes back to the same place.

(a) $x$ increases all the time; the particle never returns, so the motion is not periodic. (b) The same sawtooth shape — a slow rise followed by a sudden return — repeats again and again. The rising part crosses $x = 0$ at $t = -3, -1, 1, 3\ \text{s}$, $2\ \text{s}$ apart, so the period is $2\ \text{s}$. (c) The particle is back at $x = 0$ every $3\ \text{s}$, but the motion between those instants is different each time, so the motion as a whole never repeats. Returning to one position is not enough. (d) The curve is a sine wave crossing zero at every whole second; one full cycle, up and back down, takes $2\ \text{s}$.

Question 13.4

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion ($\omega$ is any positive constant): (a) $\sin\omega t – \cos\omega t$ (b) $\sin^3\omega t$ (c) $3\cos(\pi/4 – 2\omega t)$ (d) $\cos\omega t + \cos3\omega t + \cos5\omega t$ (e) $\exp(-\omega^2t^2)$ (f) $1 + \omega t + \omega^2t^2$

Solution. A function represents SHM only if it can be written as a single sine or cosine, $A\cos(\omega’ t + \phi)$, of one angular frequency.

(a) Following the book’s own Example 13.3,

$$\sin\omega t – \cos\omega t = \sqrt2\left(\sin\omega t\cos\tfrac\pi4 – \cos\omega t\sin\tfrac\pi4\right) = \sqrt2\,\sin\!\left(\omega t – \tfrac\pi4\right)$$

a single sine: SHM with $T = 2\pi/\omega$.

(b) The identity $\sin3\theta = 3\sin\theta – 4\sin^3\theta$ gives

$$\sin^3\omega t = \tfrac34\sin\omega t – \tfrac14\sin3\omega t$$

a mixture of two angular frequencies, $\omega$ and $3\omega$. It is not SHM. Both terms repeat after $2\pi/\omega$, so it is periodic with $T = 2\pi/\omega$.

(c) Since cosine is even, $3\cos(\pi/4 – 2\omega t) = 3\cos(2\omega t – \pi/4)$: a single cosine of angular frequency $2\omega$. It is SHM with $T = 2\pi/2\omega = \pi/\omega$.

(d) Three angular frequencies, so not SHM. The periods of the three terms are $2\pi/\omega$, $2\pi/3\omega$ and $2\pi/5\omega$; the first is a whole number of times each of the others, so all three repeat together after $2\pi/\omega$. Periodic, $T = 2\pi/\omega$.

(e) $\exp(-\omega^2t^2)$ rises to 1 at $t = 0$ and falls away to zero on both sides, never to rise again. Not periodic.

(f) $1 + \omega t + \omega^2t^2$ grows without limit. Not periodic (and no real particle could follow it for all time).

Question 13.5

A particle is in linear simple harmonic motion between two points, A and B, $10\ \text{cm}$ apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the mid-point of AB going towards A, (d) at $2\ \text{cm}$ away from B going towards A, (e) at $3\ \text{cm}$ away from A going towards B, and (f) at $4\ \text{cm}$ away from B going towards A.

Solution. Put the origin at the mid-point $O$, the centre of the motion, so that $A$ is at $x = -5\ \text{cm}$ and $B$ at $x = +5\ \text{cm}$. Two facts give every sign. The acceleration is $a = -\omega^2x$, so it always points towards $O$: positive when the particle is on $A$’s side, negative on $B$’s side, zero at $O$. The force is $F = ma$, so it always has the same sign as the acceleration. The velocity is zero at the two ends, where the particle turns round, and elsewhere its sign is simply the direction of travel.

(a) At $A$, $x = -5$: the particle is momentarily at rest, and it is pulled towards $O$, in the positive direction. (b) At $B$, $x = +5$: at rest, pulled in the negative direction. (c) At $O$ going towards $A$: moving in the negative direction, and at the centre there is no acceleration or force. (d) $2\ \text{cm}$ from $B$ is $x = +3$: moving towards $A$, and pulled towards $O$, which is also the negative direction. (e) $3\ \text{cm}$ from $A$ is $x = -2$: moving towards $B$, and pulled towards $O$, the positive direction. (f) $4\ \text{cm}$ from $B$ is $x = +1$: moving towards $A$ and pulled towards $O$, both negative.

Question 13.6

Which of the following relationships between the acceleration $a$ and the displacement $x$ of a particle involve simple harmonic motion? (a) $a = 0.7x$ (b) $a = -200x^2$ (c) $a = -10x$ (d) $a = 100x^3$

Solution. SHM needs $a = -\omega^2x$: the acceleration proportional to $x$ and opposite to it. (a) $a = 0.7x$ is proportional to $x$ but has the wrong sign: the acceleration points away from equilibrium, so a displaced particle runs off instead of coming back. (b) $a = -200x^2$ is not proportional to $x$, and since $x^2$ is never negative it pulls the particle the same way on both sides of the origin. (c) $a = -10x$ has exactly the required form, with $\omega = \sqrt{10}\ \text{rad s}^{-1}$. (d) $a = 100x^3$ is not linear in $x$ and has the wrong sign as well.

Question 13.7

The motion of a particle executing simple harmonic motion is described by the displacement function, $x(t) = A\cos(\omega t + \phi)$. If the initial ($t = 0$) position of the particle is $1\ \text{cm}$ and its initial velocity is $\omega\ \text{cm/s}$, what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi\ \text{s}^{-1}$. If instead of the cosine function, we choose the sine function to describe the SHM: $x = B\sin(\omega t + \alpha)$, what are the amplitude and initial phase of the particle with the above initial conditions.

Solution. The amplitude and phase are the two constants the initial position and velocity must fix. Differentiating, $v(t) = -A\omega\sin(\omega t + \phi)$, so at $t = 0$:

$$A\cos\phi = 1, \qquad -A\omega\sin\phi = \omega \ \Rightarrow\ A\sin\phi = -1$$

Squaring and adding gives $A^2 = 2$, so $A = \sqrt2\ \text{cm}$. Then $\cos\phi = 1/\sqrt2$ and $\sin\phi = -1/\sqrt2$, which puts $\phi$ in the fourth quadrant: $\phi = -\pi/4$, or equivalently $7\pi/4$. The value $\omega = \pi\ \text{s}^{-1}$ does not enter, because the initial velocity is given as a multiple of $\omega$ and $\omega$ cancels.

With the sine form, $v(t) = B\omega\cos(\omega t + \alpha)$, so at $t = 0$:

$$B\sin\alpha = 1, \qquad B\omega\cos\alpha = \omega \ \Rightarrow\ B\cos\alpha = 1$$

giving $B = \sqrt2\ \text{cm}$ and $\tan\alpha = 1$ with both sine and cosine positive, so $\alpha = \pi/4$. The amplitude is the same in both descriptions, as it must be; only the phase constant depends on whether we use sine or cosine.

Question 13.8

A spring balance has a scale that reads from $0$ to $50\ \text{kg}$. The length of the scale is $20\ \text{cm}$. A body suspended from this balance, when displaced and released, oscillates with a period of $0.6\ \text{s}$. What is the weight of the body?

Solution. The scale tells us the spring constant: a load of $50\ \text{kg}$, whose weight is $50\times9.8 = 490\ \text{N}$, stretches the spring by the full $0.20\ \text{m}$. So

$$k = \frac{490}{0.20} = 2450\ \text{N m}^{-1}$$

A mass on a spring oscillates with $T = 2\pi\sqrt{m/k}$, whether the spring is horizontal or vertical, so the period gives the mass:

$$m = \frac{kT^2}{4\pi^2} = \frac{2450\times0.36}{39.48} = 22.3\ \text{kg}$$

and its weight is $mg = 22.3\times9.8 = 219\ \text{N}$.

Question 13.9

A spring having a spring constant $1200\ \text{N m}^{-1}$ is mounted on a horizontal table as shown in Fig. 13.19. A mass of $3\ \text{kg}$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0\ \text{cm}$ and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Solution. The table is horizontal, so gravity plays no part and the only horizontal force is the spring’s, $F = -kx$. The mass therefore performs SHM with

$$\omega = \sqrt{\frac km} = \sqrt{\frac{1200}{3}} = 20\ \text{rad s}^{-1}$$

It is released from rest $2.0\ \text{cm}$ from equilibrium, so that is the amplitude, $A = 0.020\ \text{m}$.

(i) $\nu = \omega/2\pi = 20/2\pi = 3.18\ \text{s}^{-1}$. (ii) The acceleration is greatest at the extreme positions, where the spring is stretched most: $a_{\max} = \omega^2A = 400\times0.020 = 8.0\ \text{m s}^{-2}$. (iii) The speed is greatest at the mean position: $v_{\max} = \omega A = 20\times0.020 = 0.40\ \text{m s}^{-1}$.

Question 13.10

In Exercise 13.9, let us take the position of mass when the spring is unstretched as $x = 0$, and the direction from left to right as the positive direction of $x$-axis. Give $x$ as a function of time $t$ for the oscillating mass if at the moment we start the stopwatch ($t = 0$), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Solution. From Question 13.9, $A = 2\ \text{cm}$ and $\omega = 20\ \text{rad s}^{-1}$ in every case; only the starting point changes. The block is on the right of the spring (Fig. 13.19), so stretching the spring moves the block to the right, towards $+x$.

(a) At the mean position $x = 0$, so we need a function that is zero at $t = 0$: $x = 2\sin20t$. This describes the mass passing through $x = 0$ moving to the right; had it been moving to the left, it would be $x = -2\sin20t$. (b) At maximum stretch the mass is at $x = +2\ \text{cm}$, at rest: $x = 2\cos20t$. (c) At maximum compression it is at $x = -2\ \text{cm}$, at rest: $x = -2\cos20t$.

All three have the same amplitude, $2\ \text{cm}$, and the same angular frequency, $20\ \text{rad s}^{-1}$, because these are fixed by the spring, the mass and how far it was pulled. Written as $2\cos(20t + \phi)$, they differ only in the phase constant: $\phi = -\pi/2$, $0$ and $\pi$.

Question 13.11

Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the $x$-projection of the radius vector of the revolving particle P, in each case.

Solution. If the radius vector $OP$ makes an angle $\theta(t)$ with the positive $x$-axis, its projection on the $x$-axis is $x = r\cos\theta(t)$. So all we need is how the angle changes with time: it starts at the angle of the initial position and changes by $2\pi t/T$, increasing for anticlockwise motion and decreasing for clockwise motion.

Question 13.11 (a)

Solution. Here $r = 3\ \text{cm}$ and $T = 2\ \text{s}$, so the angular speed is $2\pi/T = \pi\ \text{rad s}^{-1}$. P starts on the negative $y$-axis, at $\theta = -\pi/2$, and goes round clockwise, so $\theta = -\pi/2 – \pi t$:

$$x = 3\cos\!\left(-\tfrac\pi2 – \pi t\right) = 3\cos\!\left(\tfrac\pi2 + \pi t\right) = -3\sin\pi t$$

A check: just after $t = 0$ the particle moves from the bottom of the circle towards the left, so $x$ should become negative, and $-3\sin\pi t$ does.

Question 13.11 (b)

Solution. Here $r = 2\ \text{m}$ and $T = 4\ \text{s}$, so the angular speed is $\pi/2\ \text{rad s}^{-1}$. P starts on the negative $x$-axis, at $\theta = \pi$, and goes round anticlockwise, so $\theta = \pi + \pi t/2$:

$$x = 2\cos\!\left(\pi + \tfrac{\pi t}2\right) = -2\cos\tfrac{\pi t}2$$

Starting on the $x$-axis, the sense of rotation does not affect the $x$-projection at all: going the other way would give $2\cos(\pi – \pi t/2)$, which is the same function.

Note on the book’s answer. The key gives $x = -2\cos(\pi t/2)$ “where $x$ is in cm”, but Fig. 13.20(b) gives the radius as $2\ \text{m}$, so the amplitude is $2\ \text{m}$ and $x$ is in metres. The form of the answer is the same.

Question 13.12

Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial ($t = 0$) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: ($x$ is in cm and $t$ is in s). (a) $x = -2\sin(3t + \pi/3)$ (b) $x = \cos(\pi/6 – t)$ (c) $x = 3\sin(2\pi t + \pi/4)$ (d) $x = 2\cos\pi t$

Solution. A particle going anticlockwise round a circle of radius $A$ with angular speed $\omega$, starting at angle $\phi$ from the positive $x$-axis, has the $x$-projection $x = A\cos(\omega t + \phi)$. So each motion has to be rewritten in exactly that form, with $A$ and $\omega$ positive; then the radius, the angular speed and the starting angle can be read off.

Question 13.12 (a)

Solution. Using $-\sin\theta = \cos(\theta + \pi/2)$,

$$x = -2\sin\!\left(3t + \tfrac\pi3\right) = 2\cos\!\left(3t + \tfrac\pi3 + \tfrac\pi2\right) = 2\cos\!\left(3t + \tfrac{5\pi}6\right)$$

At $t = 0$ this gives $x = 2\cos150^\circ = -\sqrt3$, the same as $-2\sin60^\circ$ from the original form.

Question 13.12 (b)

Solution. Cosine is an even function, so $\cos(\pi/6 – t) = \cos(t – \pi/6)$. The rewriting matters: in the form given, the argument decreases with time, which would mean clockwise rotation, while the rewritten form has it increasing, as anticlockwise rotation requires.

$$x = \cos\!\left(t – \tfrac\pi6\right)$$

Question 13.12 (c)

Solution. Using $\sin\theta = \cos(\theta – \pi/2)$,

$$x = 3\sin\!\left(2\pi t + \tfrac\pi4\right) = 3\cos\!\left(2\pi t + \tfrac\pi4 – \tfrac\pi2\right) = 3\cos\!\left(2\pi t – \tfrac\pi4\right)$$

Question 13.12 (d)

Solution. $x = 2\cos\pi t$ is already in the required form, with $\phi = 0$: the particle starts on the positive $x$-axis.

Question 13.13

Figure 13.21(a) shows a spring of force constant $k$ clamped rigidly at one end and a mass $m$ attached to its free end. A force $\mathbf F$ applied at the free end stretches the spring. Figure 13.21(b) shows the same spring with both ends free and attached to a mass $m$ at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force $\mathbf F$.

(a) What is the maximum extension of the spring in the two cases? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Question 13.13 (a)

Solution. The spring stretches until its tension balances the pull on its end. In (a) the pull is $F$ at the free end, with the wall supplying an equal pull at the other, so $kx = F$. In (b) each end is pulled by $F$ in opposite directions. The tension throughout the spring is again $F$ — exactly the situation in (a), with the second mass doing the wall’s job — so the extension is the same.

Question 13.13 (b)

Solution. In (a) the mass is on a spring of constant $k$, so $T = 2\pi\sqrt{m/k}$.

In (b), once released, the only horizontal forces on the two masses are the spring’s pulls on them, which are equal and opposite. So the centre of mass, at the midpoint of the spring, stays at rest, and each mass oscillates as though it were fixed to that midpoint by half the spring. Half a spring stretches half as much under the same tension, so its constant is $2k$. Each mass therefore oscillates with

$$T = 2\pi\sqrt{\frac{m}{2k}}$$

Question 13.14

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of $1.0\ \text{m}$. If the piston moves with simple harmonic motion with an angular frequency of $200\ \text{rad/min}$, what is its maximum speed?

Solution. The amplitude is half the stroke, $A = 0.50\ \text{m}$. In SHM the speed is greatest at the mean position, where it is $\omega A$:

$$v_{\max} = \omega A = 200\ \text{rad min}^{-1}\times0.50\ \text{m} = 100\ \text{m min}^{-1}$$

which is about $1.7\ \text{m s}^{-1}$.

Question 13.15

The acceleration due to gravity on the surface of moon is $1.7\ \text{m s}^{-2}$. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is $3.5\ \text{s}$? ($g$ on the surface of earth is $9.8\ \text{m s}^{-2}$)

Solution. The same pendulum keeps its length $L$, and $T = 2\pi\sqrt{L/g}$, so the period varies as $1/\sqrt g$:

$$T_{\text{moon}} = T_{\text{earth}}\sqrt{\frac{g_{\text{earth}}}{g_{\text{moon}}}} = 3.5\sqrt{\frac{9.8}{1.7}} = 3.5\times2.40 = 8.4\ \text{s}$$

Weaker gravity means a weaker restoring force, so the pendulum swings more slowly.

Question 13.16

A simple pendulum of length $l$ and having a bob of mass $M$ is suspended in a car. The car is moving on a circular track of radius $R$ with a uniform speed $v$. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Solution. The car is accelerating towards the centre of the track, with centripetal acceleration $v^2/R$. In the car’s frame, the bob behaves as if, besides gravity $g$ downwards, it had an acceleration $v^2/R$ directed horizontally outwards. The two combine into an effective gravity

$$g_{\text{eff}} = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2} = \sqrt{g^2 + \frac{v^4}{R^2}}$$

The pendulum hangs along $g_{\text{eff}}$, tilted outwards from the vertical, and swings about that direction exactly as an ordinary pendulum swings about the vertical, with $g_{\text{eff}}$ in place of $g$. The mass $M$ does not appear, since it cancels as it does for any pendulum.

Question 13.17

A cylindrical piece of cork of density $\rho$, base area $A$ and height $h$ floats in a liquid of density $\rho_l$. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

$$T = 2\pi\sqrt{\frac{h\rho}{\rho_l g}}$$

where $\rho$ is the density of cork. (Ignore damping due to viscosity of the liquid.)

The book prints “a cylindrical piece of cork of density of base area A”, dropping the symbol $\rho$ for the density of the cork; the formula and the last line of the question use it, so it is restored here.

Solution. When the cork floats at rest, its weight is exactly balanced by the upthrust on the part that is submerged. Push it down by a further distance $x$: an extra volume $Ax$ of liquid is displaced, and the upthrust grows by the weight of that liquid, $Ax\rho_l g$. Nothing else changes, so this extra upthrust is the net force, and it acts upwards when $x$ is downwards:

$$F = -(A\rho_l g)\,x$$

This is a restoring force proportional to the displacement, so the motion is simple harmonic, with force constant $k = A\rho_l g$. The mass of the cork is $m = Ah\rho$, so

$$T = 2\pi\sqrt{\frac mk} = 2\pi\sqrt{\frac{Ah\rho}{A\rho_l g}} = 2\pi\sqrt{\frac{h\rho}{\rho_l g}}$$

Question 13.18

One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Solution. While the pump is connected, the mercury stands higher on the suction side. Once it is removed, both ends are open to the atmosphere, so the atmospheric pressures cancel and the only unbalanced force is the weight of the extra mercury on the higher side.

Let the tube have cross-section $A$, let the mercury have density $\rho$ and total length $L$ along the tube, and suppose at some instant the level in one arm is a distance $y$ below the equilibrium level. The mercury is incompressible, so the other arm is $y$ above it, and the two levels differ by $2y$. The column of height $2y$ standing above the lower level is not balanced by anything on the other side; its weight, $(2yA)\rho g$, drives the whole column back towards equilibrium:

$$F = -(2A\rho g)\,y$$

The restoring force is proportional to the displacement, so the motion is simple harmonic, with $k = 2A\rho g$. The whole column moves together, so the moving mass is $m = AL\rho$, and

$$T = 2\pi\sqrt{\frac mk} = 2\pi\sqrt{\frac{AL\rho}{2A\rho g}} = 2\pi\sqrt{\frac{L}{2g}}$$

Common mistakes

• Question 13.1(a): calling the swimmer’s return trip periodic. The swimmer ends where the trip began, which looks like repetition. But the trip happens once; a periodic motion has to repeat itself over and over at regular intervals.
• Question 13.3(c): taking the regular returns to $x = 0$ as a period. The particle is at $x = 0$ every $3\ \text{s}$, so it is tempting to call the period $3\ \text{s}$. The motion in between differs every time, and periodicity needs the whole pattern to repeat, not one position.
• Question 13.4(c): giving the period as $2\pi/\omega$. The angular frequency inside the cosine is $2\omega$, not $\omega$, so the period is $2\pi/2\omega = \pi/\omega$. Students read the $\omega$ off the question’s wording instead of off the function.
• Question 13.6(a): choosing $a = 0.7x$ because it is proportional to $x$. Proportionality is only half the test. Without the minus sign the acceleration points away from equilibrium, and the particle runs away instead of oscillating.
• Question 13.8: using the $50\ \text{kg}$ as the mass in the formula for $k$. The spring constant is force per unit extension, so the $50\ \text{kg}$ has to be turned into a weight, $490\ \text{N}$, first. The other slip is to stop at $22.3\ \text{kg}$ when the question asks for the weight.
• Question 13.11(a): assuming anticlockwise rotation. Reference-circle problems are usually anticlockwise, and taking it so here gives $+3\sin\pi t$. The figure shows clockwise motion, which flips the sign.
• Question 13.13(b): using $k$ for each mass in Fig. 13.21(b). The midpoint of the spring stays still, so each mass is really on half the spring, and half a spring is twice as stiff. Using $k$ gives $2\pi\sqrt{m/k}$, the answer to (a), again.
• Question 13.16: adding or subtracting $v^2/R$ from $g$. The two accelerations are at right angles, one vertical and one horizontal, so they combine as vectors: $g_{\text{eff}} = \sqrt{g^2 + v^4/R^2}$, not $g \pm v^2/R$.

Practise next

• Chapter 14, Waves — the SHM of each particle of a medium, passed along from one to the next, is what a wave is.
• Chapter 3, Motion in a Plane — uniform circular motion and centripetal acceleration $v^2/R$, the basis of the reference circle in 13.11 and 13.12 and of Question 13.16.
• Chapter 4, Laws of Motion — the force analysis behind every $F = -kx$ in this exercise.