NCERT Exemplar Solutions Class 11 Physics Oscillations — Multiple Choice Questions Q1–18

NCERT Exemplar Problems, Class 11 Physics — Oscillations, MCQ I and MCQ II. All 18 questions solved.

The book numbers these questions 14.1 to 14.18; here they are Questions 1 to 18. Questions 1 to 11 are MCQ I (exactly one option is correct) and Questions 12 to 18 are MCQ II (one or more options may be correct). Three ideas carry the whole page:

• The test for simple harmonic motion. A motion is simple harmonic exactly when $a = -\omega^2 x$, that is, when the displacement can be written as a single sinusoid $x = A\sin(\omega t + \phi)$. Periodic is not enough.
• Phase. Two instants have the same phase when they are a whole number of periods apart; then displacement and velocity repeat.
• Energy. In SHM the total energy is constant and equals the maximum kinetic energy; kinetic and potential energy trade places twice per cycle.

Key insight. Before choosing “SHM”, rewrite the displacement as one sine or cosine of a single angular frequency. If a product, a power or a square of $t$ survives inside, the motion is not SHM. Questions 1, 2, 6 and 9 turn on exactly this.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

The displacement of a particle is represented by the equation $y = 3\cos\left(\dfrac{\pi}{4} – 2\omega t\right)$. The motion of the particle is

Solution. Cosine is even, so $y = 3\cos(2\omega t – \pi/4)$. This is a single sinusoid with amplitude $3$ and angular frequency $2\omega$, so the motion is simple harmonic. The angular frequency is $2\omega$, not $\omega$, so the period is

$$T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}.$$

Option (a) is the trap: it reads the angular frequency as $\omega$ and ignores the factor $2$.

Question 2

The displacement of a particle is represented by the equation $y = \sin^3\omega t$. The motion is

Solution. Use $\sin 3\theta = 3\sin\theta – 4\sin^3\theta$:

$$y = \sin^3\omega t = \frac{3\sin\omega t – \sin 3\omega t}{4}.$$

This is the sum of two sinusoids with angular frequencies $\omega$ and $3\omega$. Each repeats after $2\pi/\omega$, so $y$ is periodic with period $2\pi/\omega$. But it is not a single sinusoid, so $d^2y/dt^2$ is not $-(\text{constant})\,y$: the motion is not simple harmonic.

Question 3

The relation between acceleration and displacement of four particles are given below:

Which one of the particles is executing simple harmonic motion?

Solution. SHM needs the acceleration to be proportional to the displacement and directed opposite to it: $a = -\omega^2 x$. Only (d) has both features, with $\omega^2 = 2$. In (a) the acceleration pushes the particle further away, so it runs off instead of oscillating. In (b) and (c) the acceleration goes as $x^2$, which does not even change sign on the two sides of $x = 0$, so it cannot be a restoring force on both sides.

Question 4

Motion of an oscillating liquid column in a U-tube is

Solution. Let the column have total length $L$, cross-section $A$ and density $\rho$. If the liquid in one arm rises by $y$, the other falls by $y$, so the levels differ by $2y$. The unbalanced column of height $2y$ produces a restoring force $F = -(2y)A\rho g$, acting on the whole mass $LA\rho$. Hence

$$\ddot y = -\frac{2A\rho g}{LA\rho}\,y = -\frac{2g}{L}\,y, \qquad T = 2\pi\sqrt{\frac{L}{2g}}.$$

The acceleration is proportional to $-y$, so the motion is simple harmonic, and $\rho$ cancels because both the restoring force and the inertia are proportional to it. The period does not depend on the density.

Question 5

A particle is acted simultaneously by mutually perpendicular simple harmonic motions $x = a\cos\omega t$ and $y = a\sin\omega t$. The trajectory of motion of the particle will be

Solution. Eliminate $t$: $x^2 + y^2 = a^2\cos^2\omega t + a^2\sin^2\omega t = a^2$. The particle stays at distance $a$ from the origin, so it moves on a circle of radius $a$. This is the reverse of the familiar picture: uniform circular motion projected on two perpendicular diameters gives two SHMs of equal amplitude and a phase difference of $\pi/2$. (A circle is a special ellipse, but the specific answer is (c).)

Question 6

The displacement of a particle varies with time according to the relation $y = a\sin\omega t + b\cos\omega t$.

Solution. Put $a = R\cos\phi$ and $b = R\sin\phi$, so that $R = \sqrt{a^2+b^2}$ and $\tan\phi = b/a$. Then

$$y = R(\sin\omega t\cos\phi + \cos\omega t\sin\phi) = \sqrt{a^2+b^2}\,\sin(\omega t + \phi).$$

Two sinusoids of the same frequency always add to a single sinusoid of that frequency, so the motion is SHM. Its amplitude is $\sqrt{a^2+b^2}$, not $a+b$: the two parts are a quarter-cycle out of step, so their maxima never coincide.

Question 7

Four pendulums A, B, C and D are suspended from the same elastic support as shown in Fig. 14.1. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,

The figure shows four bobs hanging from one horizontal elastic string held between two supports; from left to right they are D (longest), C, B (shortest) and A.

Forced oscillations and resonance have been removed from the rationalised Chapter 13 and are not in the JEE Main 2026 syllabus; this question is kept for completeness.

Solution. Pendulum A, once set swinging, shakes the elastic support at its own frequency $\frac{1}{2\pi}\sqrt{g/l_A}$. The support passes this periodic force to B, C and D, so they all undergo forced oscillations at A’s frequency. The response is largest for the pendulum whose natural frequency equals the driving frequency. C has the same length as A, so the same natural frequency: it is in resonance and swings with the largest amplitude. B and D are off resonance and respond only weakly.

Question 8

Figure 14.2 shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the $x$-projection of the radius vector of the rotating particle P is

Solution. The angular speed is $\omega = 2\pi/T = 2\pi/30 = \pi/15$ rad/s. Measure the angle $\theta$ of the radius vector from the $y$-axis in the clockwise sense, which is the sense of revolution. Then P is at $\theta = 0$ when $t = 0$ and at $\theta = \omega t$ later, and its $x$-coordinate is

$$x = B\sin\theta = B\sin\left(\frac{2\pi t}{30}\right).$$

Check the start: at $t = 0$, $x = 0$, and just after, P moves to the right, so $x$ grows positive. Options (b) and (c) are both $B\cos(\pi t/15)$, which starts at $x = B$, the wrong initial position. Option (d) equals $-B\sin(\pi t/15)$, which starts at zero but moves to the left; that would be anticlockwise revolution.

Question 9

The equation of motion of a particle is $x = a\cos(\alpha t)^2$. The motion is

Solution. Read the argument as $(\alpha t)^2$: $x = a\cos(\alpha^2 t^2)$. The particle moves back and forth between $x = a$ and $x = -a$, so the motion is oscillatory. But the phase $\alpha^2t^2$ grows faster and faster, so successive swings take less and less time: the zeros come at $\alpha^2t^2 = \pi/2, 3\pi/2, \ldots$, which are not equally spaced in $t$. No fixed interval $T$ makes $x(t+T) = x(t)$ for all $t$, so the motion is not periodic.

(If the expression were meant as $a\cos^2(\alpha t) = \tfrac{a}{2}(1 + \cos 2\alpha t)$, it would be periodic and oscillatory. The book’s key, and the way the bracket is placed, take the square on $\alpha t$.)

Question 10

A particle executing S.H.M. has a maximum speed of $30$ cm/s and a maximum acceleration of $60$ cm/s². The period of oscillation is

Solution. In SHM $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$. Dividing removes the unknown amplitude:

$$\omega = \frac{a_{\max}}{v_{\max}} = \frac{60}{30} = 2 \text{ rad/s}, \qquad T = \frac{2\pi}{\omega} = \pi \text{ s}.$$

(The amplitude is $A = 30/2 = 15$ cm, though it is not needed.)

Question 11

When a mass $m$ is connected individually to two springs $S_1$ and $S_2$, the oscillation frequencies are $\nu_1$ and $\nu_2$. If the same mass is attached to the two springs as shown in Fig. 14.3, the oscillation frequency would be

The figure shows the block $m$ on a smooth horizontal floor between two walls, joined to the left wall by $S_1$ and to the right wall by $S_2$.

Solution. Alone, $\nu_1 = \frac{1}{2\pi}\sqrt{k_1/m}$ and $\nu_2 = \frac{1}{2\pi}\sqrt{k_2/m}$, so $k_1 = 4\pi^2m\nu_1^2$ and $k_2 = 4\pi^2m\nu_2^2$. Between the walls, a displacement $x$ stretches one spring by $x$ and compresses the other by $x$, and both forces point back towards equilibrium. The total restoring force is $-(k_1+k_2)x$, so the springs act in parallel:

$$\nu = \frac{1}{2\pi}\sqrt{\frac{k_1+k_2}{m}} = \sqrt{\nu_1^2+\nu_2^2}.$$

Question 12

The rotation of earth about its axis is

Solution. The Earth repeats its orientation every day, so the rotation is periodic. It is not simple harmonic: SHM is a to-and-fro motion about a mean position under a restoring force, whereas the Earth turns steadily in one sense with no mean position and no restoring torque. So (a) and (c) are both true.

Question 13

Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is

Solution. Near the bottom, a smooth bowl of radius of curvature $R$ behaves like a simple pendulum of length $R$: the component of gravity along the surface is $-mg\sin\theta \approx -mg\theta$ for a small angle $\theta$. The restoring force is proportional to the displacement and opposite to it, so the motion is simple harmonic with $T = 2\pi\sqrt{R/g}$, and every SHM is periodic. Because the ball is released only slightly above the lowest point, the small-angle condition holds. So (a) and (c) are true and (d) is false.

Question 14

Displacement vs. time curve for a particle executing S.H.M. is shown in Fig. 14.4. Choose the correct statements.

Solution. The curve crosses zero at $1, 3, 5, 7$ s and has extremes at $2, 4, 6$ s, so a full cycle takes $T = 4$ s. Two instants have the same phase exactly when they differ by a whole number of periods.

• (a) $2 – 0 = 2$ s $= T/2$. At $t = 0$ the particle is at its positive extreme and at $t = 2$ s at its negative extreme: opposite phase. False.
• (b) $6 – 2 = 4$ s $= T$. Same phase (both negative extremes). True.
• (c) $7 – 1 = 6$ s $= 1.5\,T$. Both are zero crossings, but at $t = 1$ s the particle moves downwards and at $t = 7$ s upwards. The displacement agrees but the velocity does not, so the phase differs by $\pi$. False.
• (d) $5 – 1 = 4$ s $= T$. Same phase (both crossing downwards). True.

Question 15

Which of the following statements is/are true for a simple harmonic oscillator?

Solution. (a) is the defining force law, $F = -kx$. Then $x = A\sin(\omega t + \phi)$ repeats every $2\pi/\omega$, so (b) holds, and so does its derivative $v = A\omega\cos(\omega t + \phi)$, which is (d). The acceleration $a = -\omega^2x$ changes with $x$, from zero at the mean position to $\omega^2A$ at the extremes, so (c) is false.

Question 16

The displacement time graph of a particle executing S.H.M. is shown in Fig. 14.5. Which of the following statement is/are true?

Solution. The graph is $x = A\cos(2\pi t/T)$: a maximum at $t = 0$ and $t = T$, zero at $T/4$ and $3T/4$, a minimum at $T/2$.

• (a) At $3T/4$ the displacement is zero, so the force $F = -kx$ is zero. True.
• (b) At $4T/4 = T$ the particle is at an extreme, so $|a| = \omega^2A$ is the largest it gets. True.
• (c) At $T/4$ the particle passes through the mean position, where the speed is maximum. True.
• (d) At $T/2$ the particle is at an extreme: KE $= 0$ and PE is the whole energy. They are equal only where $x = A/\sqrt2$, at $t = T/8, 3T/8, \ldots$. False.

Question 17

A body is performing S.H.M. Then its

Solution. Take $v = v_{\max}\cos\theta$ with $\theta = \omega t + \phi$ running uniformly over one cycle.

• (a) The total energy is constant, so its average is itself, and it equals the KE at the mean position, the maximum KE. True.
• (b) KE $= K_{\max}\cos^2\theta$ and the average of $\cos^2\theta$ over a cycle is $\tfrac12$. True.
• (c) The average of $\cos\theta$ over a cycle is $0$: the particle ends where it started, so its mean velocity is zero. The value $\tfrac{2}{\pi}v_{\max}$ is the mean speed, the average of $|\cos\theta|$. False.
• (d) $v_{\text{rms}} = \sqrt{\langle v^2\rangle} = v_{\max}\sqrt{\langle\cos^2\theta\rangle} = v_{\max}/\sqrt2$. True.

Question 18

A particle is in linear simple harmonic motion between two points A and B, $10$ cm apart (Fig. 14.6). Take the direction from A to B as the $+ve$ direction and choose the correct statements.

Solution. O is the mean position. The positive direction is from A towards B, and the acceleration and force always point towards O.

• (a) $3$ cm from A is the point C (since $BC = 8$ cm). Going towards B, the velocity points from A to B: positive. C lies between O and A, so “towards O” is also the A-to-B direction: acceleration and force positive. True.
• (b) At C, going towards O is going towards B, which is the positive direction. The velocity is positive, not negative. False.
• (c) $4$ cm from B is $1$ cm on the B side of O. Going towards A, the velocity is negative. The point is between B and O, so “towards O” is the B-to-A direction: acceleration and force negative. True.
• (d) At B the particle is momentarily at rest, but the acceleration and force point towards O, from B to A: negative. True.

Common mistakes

• Reading the angular frequency off the bracket in Question 1. In $\cos(\pi/4 – 2\omega t)$ the angular frequency is $2\omega$, so $T = \pi/\omega$.
• Calling every periodic motion SHM in Questions 2, 12 and 13. $\sin^3\omega t$ and the Earth’s rotation are periodic but not simple harmonic.
• Adding amplitudes in Question 6. Sinusoids a quarter-cycle apart combine as $\sqrt{a^2+b^2}$, like perpendicular vectors.
• Ignoring the sense of revolution in Question 8. Clockwise from the top gives $+B\sin\omega t$; anticlockwise would give $-B\sin\omega t$.
• Matching displacement only in Question 14. Same phase needs the same displacement and the same direction of motion.
• Confusing mean velocity with mean speed in Question 17. Over a cycle the mean velocity is zero; $\tfrac{2}{\pi}v_{\max}$ is the mean speed.
• Taking “towards O” as always negative in Question 18. Its sign depends on which side of O the particle is.

Practise next

• Problem. A particle in SHM has a maximum speed of $20$ cm/s and a maximum acceleration of $80$ cm/s². Find its period and amplitude.

Show answer

$T = \pi/2$ s, $A = 5$ cm. Dividing $a_{\max} = A\omega^2$ by $v_{\max} = A\omega$ gives $\omega = 80/20 = 4$ rad/s, so $T = 2\pi/4 = \pi/2$ s, and then $A = v_{\max}/\omega = 20/4 = 5$ cm, the same method as Question 10.

• Problem. Is the motion $y = \sin^2\omega t$ simple harmonic? If so, find its mean position, amplitude and period.

Show answer

Yes: SHM about $y = \tfrac12$, amplitude $\tfrac12$, period $\pi/\omega$. Since $\sin^2\omega t = \tfrac12 – \tfrac12\cos 2\omega t$, it is a single sinusoid of angular frequency $2\omega$ shifted to the mean position $\tfrac12$. Unlike $\sin^3\omega t$ in Question 2, no second frequency appears.

• Exemplar Oscillations: Very Short, Short and Long Answer Questions — where the same ideas are applied to springs, pulleys, floating logs, a tunnel through the Earth and a pendulum whose string snaps.
• 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.