NCERT Exemplar Solutions Class 11 Physics Oscillations — Very Short, Short and Long Answer Questions Q19–40

NCERT Exemplar Problems, Class 11 Physics — Oscillations, Very Short Answer Questions 19 to 28, Short Answer Questions 29 to 34 and Long Answer Questions 35 to 40. All 22 questions solved. The book numbers them 14.19 to 14.40.

Three ideas carry this page. Find the restoring force for a small displacement: if it comes out as $F = -kx$, the motion is SHM with $\omega = \sqrt{k/m}$, whether the “spring” is a real spring, buoyancy, gravity inside the Earth or a column of liquid. Phase locates the oscillator in its cycle: $x = A\sin(\omega t + \phi)$, with velocity a quarter-cycle ahead of displacement. Energy is shared, not lost: $U = \tfrac12kx^2$ and $K = \tfrac12k(A^2 – x^2)$ add to a constant.

Key insight. In every long answer below, the work is the same two lines: displace the system by a small $x$ from equilibrium, write the net force, and read off $k_{\text{eff}}$ and the inertia. The weight or the equilibrium buoyancy always cancels; only the extra force from the displacement matters. That is also where the printed key goes wrong in Questions 38 and 40.

Very Short Answer Questions

Question 19

Displacement versus time curve for a particle executing S.H.M. is shown in Fig. 14.7. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.

Solution. Velocity is the slope of the displacement–time graph. It is zero at the crests and troughs, where the particle is at an extreme and turns round: these are A ($t = 0$, a maximum), C (minimum), E (maximum) and G (minimum). The speed is greatest where the graph is steepest, at the zero crossings, where the particle passes through the mean position: B, D, F and H.

Question 20

Two identical springs of spring constant $K$ are attached to a block of mass $m$ and to fixed supports as shown in Fig. 14.8. When the mass is displaced from equilibrium position by a distance $x$ towards right, find the restoring force.

The figure shows the block on a smooth floor inside a frame, joined to the left wall by one spring and to the right wall by the other.

Solution. When the block moves $x$ to the right, the left spring is stretched by $x$ and pulls it to the left with force $Kx$. The right spring is compressed by $x$ and pushes it to the left with force $Kx$. Both forces point back towards equilibrium, so they add:

$$F = -2Kx.$$

Question 21

What are the two basic characteristics of a simple harmonic motion?

Solution. The defining equation is $a = -\omega^2 x$. It says two things at once: the size of the acceleration (and of the restoring force) is proportional to the displacement from the mean position, and its direction is always opposite to the displacement, that is, towards the mean position.

Question 22

When will the motion of a simple pendulum be simple harmonic?

Solution. The restoring force along the arc is $-mg\sin\theta$, which is proportional to $\sin\theta$, not to $\theta$. Only when the angular displacement is small does $\sin\theta \approx \theta$ (in radians), so that $F \approx -mg\theta = -(mg/l)s$ is proportional to the arc displacement $s$. The motion is then SHM with $T = 2\pi\sqrt{l/g}$.

Question 23

What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

Solution. $a_{\max} = \omega^2 A$ and $v_{\max} = \omega A$, so

$$\frac{a_{\max}}{v_{\max}} = \omega.$$

Question 24

What is the ratio between the distance travelled by the oscillator in one time period and amplitude?

Solution. In one period the oscillator goes from the mean position to one extreme ($A$), back to the mean ($A$), to the other extreme ($A$) and back ($A$): a distance of $4A$. The ratio is $4A/A = 4$. (The displacement over one period is zero.)

Question 25

In Fig. 14.9, what will be the sign of the velocity of the point P′, which is the projection of the velocity of the reference particle P? P is moving in a circle of radius $R$ in anticlockwise direction.

Solution. P is in the first quadrant at angle $\omega t + \phi$ and moves anticlockwise, so its velocity, of size $R\omega$, is tangent to the circle and points up and to the left. Its $x$-component is $-R\omega\sin(\omega t + \phi)$, which is negative in the first quadrant. P′ is therefore moving towards O, in the $-x$ direction.

Question 26

Show that for a particle executing S.H.M, velocity and displacement have a phase difference of $\pi/2$.

Solution. Let $x = A\sin(\omega t + \phi)$. Differentiating,

$$v = \frac{dx}{dt} = A\omega\cos(\omega t + \phi) = A\omega\sin\left(\omega t + \phi + \frac{\pi}{2}\right).$$

The phase of $v$ is $\omega t + \phi + \pi/2$ and the phase of $x$ is $\omega t + \phi$, so velocity leads displacement by $\pi/2$ at every instant. This is why $v = 0$ when $|x|$ is largest and $|v|$ is largest when $x = 0$.

Question 27

Draw a graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.

Solution. With $x$ the displacement and $A$ the amplitude,

$$U = \tfrac12 kx^2, \qquad K = \tfrac12 k(A^2 – x^2), \qquad E = K + U = \tfrac12 kA^2.$$

$U$ is an upward parabola, zero at $x = 0$ and equal to $E$ at $x = \pm A$. $K$ is the same parabola turned upside down, equal to $E$ at $x = 0$ and zero at $x = \pm A$. The total energy is a horizontal line at $E$. The two curves cross at $x = \pm A/\sqrt2$, where $K = U = E/2$.

Question 28

The length of a second’s pendulum on the surface of Earth is $1$ m. What will be the length of a second’s pendulum on the moon?

Solution. A second’s pendulum has the same period ($2$ s) wherever it is, so $T = 2\pi\sqrt{l/g}$ fixed means $l \propto g$. On the Moon $g_M = g_E/6$, so

$$l_M = \frac{g_M}{g_E}\,l_E = \frac{1}{6}\times 1 \text{ m} \approx 0.167 \text{ m}.$$

Short Answer Questions

Question 29

Find the time period of mass $M$ when displaced from its equilibrium position and then released for the system shown in Fig 14.10.

The figure shows a light pulley with the mass $M$ hanging from its axle. An inextensible string passes under the pulley; one end of the string is tied to the ceiling and the other end is attached to a vertical spring (spring constant $k$) whose top is fixed to the ceiling.

Solution. The tension $T$ is the same in the string and the spring, and two strands hold up the pulley. If the pulley (with $M$) moves down by $y$, each strand must lengthen by $y$. The string cannot stretch, so the spring takes up the whole extra length: it stretches by $2y$.

In equilibrium the spring is already stretched by $y_0$ with $2ky_0 = Mg$. Now pull $M$ down by a further $x$. The spring stretches by an extra $2x$, the tension becomes $k(y_0 + 2x)$, and the upward pull of the two strands is $2k(y_0 + 2x)$. The net force on $M$ is

$$F = Mg – 2k(y_0 + 2x) = -4kx.$$

So the system behaves as a mass $M$ on a spring of constant $4k$:

$$T = 2\pi\sqrt{\frac{M}{4k}} = \pi\sqrt{\frac{M}{k}}.$$

Question 30

Show that the motion of a particle represented by $y = \sin\omega t – \cos\omega t$ is simple harmonic with a period of $2\pi/\omega$.

Solution. Take out $\sqrt2$:

$$y = \sqrt2\left(\frac{1}{\sqrt2}\sin\omega t – \frac{1}{\sqrt2}\cos\omega t\right) = \sqrt2\left(\sin\omega t\cos\frac{\pi}{4} – \cos\omega t\sin\frac{\pi}{4}\right) = \sqrt2\sin\left(\omega t – \frac{\pi}{4}\right).$$

This is a single sinusoid, so $d^2y/dt^2 = -\omega^2 y$: the motion is SHM with amplitude $\sqrt2$, initial phase $-\pi/4$ and period $2\pi/\omega$.

Question 31

Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.

Solution. The total (maximum) energy is $E = \tfrac12 kA^2$. Setting the potential energy to half of it:

$$\tfrac12 kx^2 = \tfrac12\cdot\tfrac12 kA^2 \implies x^2 = \frac{A^2}{2} \implies x = \pm\frac{A}{\sqrt2}.$$

At this point the kinetic energy is the other half, so this is also where $K = U$.

Question 32

A body of mass $m$ is situated in a potential field $U(x) = U_0(1 – \cos\alpha x)$ when $U_0$ and $\alpha$ are constants. Find the time period of small oscillations.

Solution. The force is

$$F = -\frac{dU}{dx} = -U_0\alpha\sin\alpha x.$$

For small oscillations $\alpha x$ is small, so $\sin\alpha x \approx \alpha x$ and $F \approx -U_0\alpha^2 x$. This has the form $F = -kx$ with $k = U_0\alpha^2$. (Equivalently, $U \approx \tfrac12 U_0\alpha^2x^2$ near $x = 0$.) Hence

$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{m}{U_0\alpha^2}}.$$

Question 33

A mass of $2$ kg is attached to the spring of spring constant $50$ N m⁻¹. The block is pulled to a distance of $5$ cm from its equilibrium position at $x = 0$ on a horizontal frictionless surface from rest at $t = 0$. Write the expression for its displacement at anytime $t$.

Solution. $\omega = \sqrt{k/m} = \sqrt{50/2} = 5$ rad/s and the amplitude is $5$ cm. Write $x = A\sin(\omega t + \phi)$. At $t = 0$ the block is at $x = 5$ cm and at rest, so $\sin\phi = 1$ and $\phi = \pi/2$:

$$x = 5\sin\left(5t + \frac{\pi}{2}\right) = 5\cos 5t \text{ cm}.$$

Check: $x(0) = 5$ cm and $v(0) = -25\sin 0 = 0$, as required.

Printed key error. The key gives $x = 5\sin 5t$. That has $x(0) = 0$ and $v(0) = 25$ cm/s, which describes a block pushed from the equilibrium position, not one released from rest at $5$ cm. The correct form is $x = 5\cos 5t$.

Question 34

Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^\circ$ to the right with the vertical, the other pendulum makes an angle of $1^\circ$ to the left of the vertical. What is the phase difference between the pendulums?

Solution. Take right as positive and write $\theta_1 = \theta_0\sin(\omega t + \delta_1)$ and $\theta_2 = \theta_0\sin(\omega t + \delta_2)$ with $\theta_0 = 2^\circ$. At the instant described:

• the first is at its right extreme, so $\sin(\omega t + \delta_1) = 1$ and $\omega t + \delta_1 = 90^\circ$;
• the second is at $-1^\circ$, so $\sin(\omega t + \delta_2) = -\tfrac12$ and $\omega t + \delta_2 = -30^\circ$ (here $\cos(-30^\circ) > 0$, so it is swinging back towards the vertical).

Subtracting, $\delta_1 – \delta_2 = 120^\circ$. (Had the second been at $-1^\circ$ on its way out to the left, the phase would be $210^\circ$, and the difference $210^\circ – 90^\circ$ is again $120^\circ$.)

Long Answer Questions

Question 35

A person normally weighing $50$ kg stands on a massless platform which oscillates up and down harmonically at a frequency of $2.0$ s⁻¹ and an amplitude $5.0$ cm. A weighing machine on the platform gives the persons weight against time.

(a) Will there be any change in weight of the body, during the oscillation?

(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?

Solution. (a) Yes. The machine reads the normal force $N$, not $mg$. Taking upwards as positive, $N – mg = ma$, so $N = m(g + a)$, and $a$ changes through the cycle.

(b) $\omega = 2\pi\nu = 4\pi$ rad/s, so $\omega^2A = 16\pi^2 \times 0.05 = 7.90$ m/s². The acceleration is $a = -\omega^2 y$, largest in size at the extremes.

• At the lowest point ($y = -A$) the acceleration is $+\omega^2A$, upwards: $$N_{\max} = m(g + \omega^2A) = 50(9.8 + 7.90) \approx 885 \text{ N}.$$
• At the highest point ($y = +A$) the acceleration is $-\omega^2A$, downwards: $$N_{\min} = m(g – \omega^2A) = 50(9.8 – 7.90) \approx 95 \text{ N}.$$

The machine reads about $885$ N (about $90$ kg-wt) at the bottom and about $95$ N (about $9.7$ kg-wt) at the top, and the true $490$ N as the platform passes the mean position.

Printed key error. The key rounds $50 \times 0.05 \times 16\pi^2 = 394.8$ N to $400$ N and so gives $890$ N and $90$ N, which is only rounding. But it then says the maximum is at the topmost position and the minimum at the lowermost. That is the wrong way round: at the top the platform is accelerating downwards, so it presses on the person less. The maximum is at the lowest point and the minimum at the highest.

Question 36

A body of mass $m$ is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is $4$ cm below the point, where it was held in hand.

(a) What is the amplitude of oscillation?

(b) Find the frequency of oscillation?

Solution. (a) The mass starts from rest, so the release point is an extreme position. The lowest point, $4$ cm below, is the other extreme. The mean position lies midway, so the amplitude is $A = 4/2 = 2$ cm.

(Energy gives the same: falling $4$ cm with zero speed at both ends, $mg(0.04) = \tfrac12k(0.04)^2$, so $k = mg/0.02$, and the equilibrium extension $mg/k$ is $2$ cm.)

(b) At the mean position $kx_0 = mg$ with $x_0 = 0.02$ m, so

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{x_0}} = \sqrt{\frac{9.8}{0.02}} = 22.1 \text{ rad/s}, \qquad \nu = \frac{\omega}{2\pi} \approx 3.5 \text{ Hz}.$$

Printed key error. The key gives the frequency as $2.8$ s⁻¹. With $x_0 = 2$ cm (its own amplitude), $\nu = \frac{1}{2\pi}\sqrt{9.8/0.02} = 3.52$ Hz. The period is $T = 1/\nu = 0.28$ s, which suggests the key printed the period’s digits as the frequency.

Question 37

A cylindrical log of wood of height $h$ and area of cross-section $A$ floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period

$$T = 2\pi\sqrt{\frac{m}{A\rho g}}$$

where $m$ is mass of the body and $\rho$ is density of the liquid.

Solution. Let the log float upright with depth $x_0$ submerged. In equilibrium the buoyant force balances the weight: $A x_0\rho g = mg$.

Push it down by a further $x$. The submerged depth becomes $x_0 + x$, and the buoyant force $A(x_0 + x)\rho g$ now exceeds the weight. The net force, taking downwards as positive, is

$$F = mg – A(x_0 + x)\rho g = -(A\rho g)\,x.$$

The force is proportional to the displacement and opposite to it, so the motion is SHM with $k = A\rho g$, and

$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{m}{A\rho g}}.$$

(This neglects the mass of water dragged along and any viscous damping, and assumes the log stays partly submerged and does not tilt.)

Question 38

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of $45^\circ$ each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Solution. Let the tube have cross-section $A$, let the mercury have density $\rho$, and let the length of mercury along each arm be $l$ in equilibrium, so the whole column has length $2l$ and mass $2lA\rho$.

Suppose the mercury moves a distance $y$ along the tube, so that the length in one arm becomes $l + y$ and in the other $l – y$. Measured vertically, one surface rises by $y\sin45^\circ$ and the other falls by $y\sin45^\circ$, so the surfaces differ in height by $2y\sin45^\circ$. That unbalanced column exerts a pressure $\rho g(2y\sin45^\circ)$, which acts on the area $A$ and drives the whole column back:

$$F = -2A\rho g\sin45^\circ\,y.$$

Newton’s second law for the whole column gives

$$2lA\rho\,\ddot y = -2A\rho g\sin45^\circ\,y \implies \ddot y = -\frac{g\sin45^\circ}{l}\,y.$$

The acceleration is proportional to $-y$, so yes, the motion is simple harmonic, with

$$T = 2\pi\sqrt{\frac{l}{g\sin45^\circ}} = 2\pi\sqrt{\frac{\sqrt2\,l}{g}}.$$

Energy check: an arm holding length $s$ of mercury has its centre of mass at height $\tfrac12 s\sin45^\circ$, so its PE is $\tfrac12A\rho g\sin45^\circ\,s^2$. The total PE is $\tfrac12A\rho g\sin45^\circ[(l+y)^2 + (l-y)^2]$, which rises by $A\rho g\sin45^\circ\,y^2$, while the KE is $\tfrac12(2lA\rho)\dot y^2$. Constant total energy then gives the same $\ddot y = -(g\sin45^\circ/l)\,y$.

Printed key error. The key works by energy but takes the mass of a slice as $A\rho\,dx$ with $dx$ measured vertically. In a tube inclined at $45^\circ$, a vertical rise $dx$ corresponds to a length $dx/\sin45^\circ$ of tube, so it undercounts the mass. It then drops the $\sin^2 45^\circ$ factor between lines and ends with $T = 2\pi\sqrt{l/g}$. The correct period is $2\pi\sqrt{l/(g\sin45^\circ)}$, which is longer by a factor $2^{1/4} \approx 1.19$. As a check, setting the angle to $90^\circ$ (a U-tube with vertical arms) gives $T = 2\pi\sqrt{l/g} = 2\pi\sqrt{L/2g}$ with $L = 2l$, the standard U-tube result.

Question 39

A tunnel is dug through the centre of the Earth. Show that a body of mass $m$ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

Solution. Treat the Earth as a uniform sphere of radius $R$ and density $\rho$. At distance $x$ from the centre, only the mass inside radius $x$ attracts the body (the shell outside exerts no net force), so

$$g(x) = \frac{G\cdot\frac43\pi x^3\rho}{x^2} = \frac{4}{3}\pi G\rho\,x = \frac{g}{R}\,x,$$

using $g = \tfrac43\pi G\rho R$ at the surface. The force on the body points towards the centre:

$$F = -\frac{mg}{R}\,x.$$

This is $F = -kx$ with $k = mg/R$, so the body executes SHM about the centre, with amplitude $R$ (it is dropped from rest at the surface) and period

$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{R}{g}} \approx 2\pi\sqrt{\frac{6.4\times10^6}{9.8}} \text{ s} \approx 84.6 \text{ min}.$$

Question 40

A simple pendulum of time period $1$ s and length $l$ is hung from a fixed support at O, such that the bob is at a distance $H$ vertically above A on the ground (Fig. 14.11). The amplitude is $\theta_0$. The string snaps at $\theta = \theta_0/2$. Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume $\theta_0$ to be small so that $\sin\theta_0 \approx \theta_0$ and $\cos\theta_0 \approx 1$.

The figure shows the pendulum OP of length $l$ swinging between angles $\pm\theta_0$; at the mean position the bob P is a height $H$ directly above the point A on the ground.

Solution. Take $t = 0$ at the extreme $\theta = \theta_0$, so $\theta = \theta_0\cos\omega t$ with $\omega = 2\pi/T = 2\pi$ rad/s.

When and how fast it snaps. $\theta = \theta_0/2$ first when $\cos 2\pi t_1 = \tfrac12$, at $t_1 = \tfrac16$ s. Then

$$\dot\theta = -2\pi\theta_0\sin\frac{2\pi}{6} = -\sqrt3\,\pi\theta_0,$$

so the bob’s speed is $u = \sqrt3\,\pi\theta_0\,l$, perpendicular to the string and directed towards the vertical through O. Its vertical component is $u\sin(\theta_0/2)$, of order $\theta_0^2$, so to first order in $\theta_0$ the velocity is horizontal.

Time to fall. At the snap, the bob is $l(1 – \cos\tfrac{\theta_0}{2}) \approx 0$ above its lowest position, so it falls a height $H$ from rest vertically:

$$t = \sqrt{\frac{2H}{g}}.$$

Where it lands. At the snap the bob is a horizontal distance $l\sin\tfrac{\theta_0}{2} \approx \tfrac12 l\theta_0$ from the vertical through A. During the fall it moves horizontally towards that vertical by

$$ut = \sqrt3\,\pi\theta_0 l\sqrt{\frac{2H}{g}} = \pi\theta_0 l\sqrt{\frac{6H}{g}}.$$

Its distance from A, measured on the side where the string snapped, is therefore

$$X = \frac{l\theta_0}{2} – \pi l\theta_0\sqrt{\frac{6H}{g}} = l\theta_0\left(\frac12 – \pi\sqrt{\frac{6H}{g}}\right).$$

If this comes out negative, the bob lands on the other side of A, at distance $l\theta_0\left(\pi\sqrt{6H/g} – \tfrac12\right)$. Because the period is $1$ s, $l = g/4\pi^2 \approx 0.25$ m.

Printed key error. The key sets up the motion the same way, but makes two slips at the end. It takes the bob’s horizontal distance from A at the moment of snapping as $l\sin\theta_0 = l\theta_0$, whereas the string snaps at $\theta_0/2$, so the distance is $l\theta_0/2$. And it writes $\sqrt3\,\pi\theta_0 l\sqrt{2H/g}$ as $\sqrt{6H/g}\,\theta_0 l$, losing the factor $\pi$. Its final $l\theta_0(1 – \sqrt{6H/g})$ should be $l\theta_0\left(\tfrac12 – \pi\sqrt{6H/g}\right)$.

Common mistakes

• Counting only one spring in Question 20. On the stretched side one spring pulls and on the other side one pushes; both point the same way, giving $2Kx$.
• Taking the spring stretch equal to the pulley’s drop in Question 29. The string is inextensible, so the spring takes up both strands’ extra length: $2y$, and the effective constant is $4k$.
• Using $\sin$ for a block released from rest at an extreme in Question 33. Released from rest at $x = A$ means $x = A\cos\omega t$.
• Putting the heaviest reading at the top in Question 35. At the top the acceleration is downwards and the normal force is smallest.
• Treating the release point as the mean position in Question 36. It is an extreme; the mean is halfway to the lowest point.
• Measuring the mercury’s mass along the vertical in Question 38. Mass goes with length along the inclined tube; only the restoring head depends on vertical height.
• Using the amplitude $\theta_0$ for the snapping position in Question 40. The string snaps at $\theta_0/2$, so the bob starts $l\theta_0/2$ from the vertical.

Practise next

• Problem. A mass hangs from a light vertical spring and is released from rest with the spring unstretched. It falls a greatest distance of $10$ cm. Find the amplitude and the frequency of the oscillation ($g = 9.8$ m/s²).

Show answer

$A = 5$ cm, $\nu \approx 2.2$ Hz. The release point and the lowest point are the two extremes, so the mean is $5$ cm below the release point and $mg/k = 0.05$ m. Then $\omega = \sqrt{g/0.05} = 14.0$ rad/s and $\nu = \omega/2\pi \approx 2.23$ Hz, by the method of Question 36.

• Problem. A platform oscillates vertically in SHM at $2$ Hz. What is the smallest amplitude at which a block resting on it just loses contact, and where does that first happen?

Show answer

About $6.2$ cm, at the top of the motion. Contact is lost when the downward acceleration at the top reaches $g$: $\omega^2A = g$ with $\omega = 4\pi$ rad/s gives $A = 9.8/(16\pi^2) = 0.062$ m. At the top the normal force $m(g – \omega^2A)$ is smallest, as in Question 35.

• Problem. A V-tube has both arms inclined at $60^\circ$ to the horizontal and holds liquid with $20$ cm along each arm. Find the period of small oscillations ($g = 9.8$ m/s²).

Show answer

About $0.96$ s. As in Question 38, $T = 2\pi\sqrt{l/(g\sin60^\circ)} = 2\pi\sqrt{0.20/(9.8\times0.866)} \approx 0.965$ s.

• Exemplar Oscillations: Multiple Choice Questions — the same ideas tested as quick options on phase, period, springs and energy.
• NCERT textbook Chapter 13 — Oscillations for the theory of SHM, its link with uniform circular motion, energy and the simple pendulum.
• Exemplar Kinetic Theory — the previous unit.
• Exemplar Waves — the next unit, where an oscillation travels through a medium.