Exercises — Oscillations

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13.12 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.1
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.
Hint. Periodic motion must repeat itself at regular intervals — a single, non-repeated event does not qualify, no matter how it looks.
(a) NOT periodic — this describes a single completed trip, not a motion that keeps repeating itself at regular intervals.

(b) Periodic — the magnet oscillates back and forth about the N-S direction under a restoring torque, repeating regularly.

(c) Periodic — uniform rotation about the centre of mass repeats identically every revolution, even though it is not oscillatory.

(d) NOT periodic — once released, the arrow flies away and never returns to repeat its motion.

✦ (b) and (c) are periodic since they repeat regularly, while (a) and (d) each describe a single non-repeating event, which is why they fail the basic definition regardless of how motion-like they appear.
13.22 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.2
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 lowermost point. (d) general vibrations of a polyatomic molecule about its equilibrium position.
Hint. SHM specifically needs a linear restoring force pulling back toward one mean position — uniform rotation has no such restoring force at all, and a molecule's vibration is usually many different SHM-like modes added together, not one.
(a) Periodic, but NOT SHM — uniform rotation has no restoring force pulling it back to a mean position; it just repeats due to its own uniform angular motion.

(b) Nearly SHM — the pressure imbalance between the two mercury columns provides a restoring force proportional to displacement (this is exactly what Q13.18 asks you to prove).

(c) Nearly SHM — for small displacements near the lowest point, gravity provides a restoring force proportional to displacement, just like a pendulum.

(d) Periodic, but generally NOT SHM — a polyatomic molecule's vibration is usually a superposition of several different normal-mode frequencies, not one single sinusoid.

✦ (b) and (c) qualify as SHM because each has a genuine single restoring force linear in displacement, while (a) has no restoring force at all and (d) typically combines multiple frequencies at once.
13.33 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.3
Fig. 13.18 depicts four x-t plots for linear motion of a particle: (a) a curve rising monotonically with time, never repeating; (b) a jagged waveform repeating with peaks at t = -3, -1, 1, 3 s; (c) an irregular waveform over t = 1 to 13 s whose shape does not repeat exactly from one stretch to the next; (d) a smooth wave-like curve with peaks spaced 2 s apart. Which of the plots represent periodic motion? What is the period of motion in each periodic case?
Hint. A plot is periodic only if its exact shape repeats after some fixed interval — an irregular pattern that looks vaguely wave-like but never repeats identically is not periodic.
(a) NOT periodic — the curve keeps rising and never returns to repeat a previous value.

(b) Periodic, with period T = 2 s — the jagged pattern repeats identically every 2 seconds (peaks at -3, -1, 1, 3 s confirm the 2 s spacing).

(c) NOT periodic — although it looks wave-like, the exact shape between t = 1-7 s differs from the shape between t = 7-13 s (a tall spike appears in the second stretch that has no counterpart in the first), so it never exactly repeats.

(d) Periodic, with period T = 2 s — the smooth curve repeats identically every 2 seconds.

✦ Only (b) and (d) are periodic, each with T = 2 s, since periodicity requires the EXACT shape to repeat, not just a generally similar-looking pattern — which is exactly why (c) fails despite looking periodic at a glance.
13.44 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.4
Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give the 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) 3cos(pi/4 - 2 omega t), (d) cos(omega t) + cos(3 omega t) + cos(5 omega t), (e) exp(-omega^2 t^2), (f) 1 + omega t + omega^2 t^2.
Hint. Rewrite each as a single sinusoid if possible using trig identities; a sum of DIFFERENT frequencies is periodic but not SHM, and an exponential or polynomial in t never repeats a value.
(i) sin(omega t) - cos(omega t) = sqrt(2) sin(omega t - pi/4) — SHM, period T = 2 pi/omega.

(ii) sin^3(omega t) = (3 sin(omega t) - sin(3 omega t))/4 — a sum of two DIFFERENT frequencies (omega and 3 omega), so periodic but NOT SHM; period T = 2 pi/omega (the larger of the two component periods).

(iii) 3cos(pi/4 - 2 omega t) = 3cos(2 omega t - pi/4) since cosine is even — SHM, period T = 2 pi/(2 omega) = pi/omega.

(iv) cos(omega t) + cos(3 omega t) + cos(5 omega t) — a sum of three different frequencies, so periodic but NOT SHM; period T = 2 pi/omega.

(v) exp(-omega^2 t^2) — a Gaussian curve that decays toward zero as t goes to plus or minus infinity and never repeats — NON-periodic.

(vi) 1 + omega t + omega^2 t^2 — a polynomial that grows without bound — NON-periodic.

✦ Only (i) and (iii) are single sinusoids and therefore true SHM; (ii) and (iv) are periodic sums of multiple frequencies, which is a fundamentally different (and non-SHM) category even though both repeat regularly.
13.54 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.5
A particle is in linear simple harmonic motion between two points, A and B, 10 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 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A.
Hint. The midpoint of AB is the mean position (the amplitude is 5 cm on either side). Acceleration and force always point toward the mean position; velocity is zero only at the two ends.
The midpoint O of AB is the mean position, with amplitude 5 cm on either side (A is the negative extreme, B the positive extreme, taking A-to-B as positive).

(a) At A (negative extreme): velocity = 0, acceleration = positive (toward mean, i.e. toward B), force = positive.

(b) At B (positive extreme): velocity = 0, acceleration = negative (toward mean, i.e. toward A), force = negative.

(c) At midpoint (mean position) going towards A: velocity = negative (moving toward A), acceleration = 0, force = 0.

(d) At 2 cm from B going towards A: position is +3 cm from mean, still moving toward A, so velocity = negative; acceleration = negative (restoring toward mean); force = negative.

(e) At 3 cm from A going towards B: position is -2 cm from mean, moving toward B, so velocity = positive; acceleration = positive (restoring toward mean); force = positive.

(f) At 4 cm from B going towards A: position is +1 cm from mean, moving toward A, so velocity = negative; acceleration = negative; force = negative.

✦ In every case acceleration and force share the same sign and always point back toward the mean position (never toward whichever end the particle is closer to), since a = -omega^2 x depends only on which side of the mean position the particle currently sits on, not on which direction it happens to be moving.
13.62 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.6
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.
Hint. SHM requires a to be LINEAR in x with a NEGATIVE proportionality constant — check both conditions, not just linearity.
(a) a = 0.7x is linear in x, but the constant is POSITIVE — this pushes the particle further away from x = 0 rather than restoring it, so it is NOT SHM.

(b) a = -200x^2 depends on x^2, not x — NOT SHM.

(c) a = -10x is linear in x with a negative constant — this IS SHM, with omega^2 = 10, so omega = sqrt(10) rad/s.

(d) a = 100x^3 depends on x^3, not x — NOT SHM.

✦ Only (c) represents SHM, since it is the only option that is both linear in x AND has a negative coefficient — linearity alone is not enough, as (a) shows.
13.74 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.7
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 cm and its initial velocity is omega cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is pi per second. 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?
Hint. Write x(0) and v(0) as two simultaneous equations in A and phi (or B and alpha), then combine them using sin-squared plus cos-squared equals one.
Cosine form: x(0) = A cos(phi) = 1. v(0) = -omega A sin(phi) = omega, so A sin(phi) = -1.

Squaring and adding: A^2 = (A cos phi)^2 + (A sin phi)^2 = 1 + 1 = 2, so A = sqrt(2) cm.

tan(phi) = sin(phi)/cos(phi) = -1/1 = -1, and since cos(phi) is positive while sin(phi) is negative, phi is in the fourth quadrant: phi = -pi/4.

Sine form: x(0) = B sin(alpha) = 1. v(0) = B omega cos(alpha) = omega, so B cos(alpha) = 1.

Squaring and adding: B^2 = 1 + 1 = 2, so B = sqrt(2) cm. Both sin(alpha) and cos(alpha) are positive, so alpha is in the first quadrant: alpha = pi/4.

✦ Both descriptions give the same amplitude, sqrt(2) cm, since they describe the identical physical motion — only the phase constant differs (-pi/4 for cosine form versus +pi/4 for sine form), exactly the pi/2 shift that relates cosine to sine.
13.83 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.8
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Hint. Use the full-scale reading and scale length to find the spring constant first, then use the oscillation period to find the mass actually hanging on it.
Step 1 — Spring constant from the scale. Full-scale force = 50 kg x 9.8 m/s^2 = 490 N, stretching the spring by the full scale length 0.20 m. k = F/x = 490/0.20 = 2450 N/m.

Step 2 — Mass from the period. T = 2 pi sqrt(m/k), so m = k(T/2pi)^2 = 2450 x (0.6/6.2832)^2 = 2450 x 0.009119 = 22.34 kg.

Step 3 — Weight. Weight = mg = 22.34 x 9.8 = 219 N.

✦ The body weighs about 219 N, since the balance's full-scale calibration (50 kg over 20 cm) is what fixes k, and the observed oscillation period then reveals the actual mass hanging on it independently of the scale reading itself.
13.93 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.9
A spring having a spring constant 1200 N/m is mounted on a horizontal table. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine (i) the frequency of oscillations, (ii) the maximum acceleration of the mass, and (iii) the maximum speed of the mass.
Hint. Find omega from k and m first; the pulled distance is the amplitude, needed for the maximum acceleration and speed formulas.
Step 1 — Angular frequency. omega = sqrt(k/m) = sqrt(1200/3) = sqrt(400) = 20 rad/s.

(i) Frequency: nu = omega/(2 pi) = 20/6.2832 = 3.18 Hz.

(ii) Maximum acceleration: a_max = omega^2 A = 400 x 0.02 = 8 m/s^2.

(iii) Maximum speed: v_max = omega A = 20 x 0.02 = 0.4 m/s.

✦ All three answers follow directly once omega is known from k and m, since amplitude alone (from how far the mass was pulled) then fixes both the maximum acceleration and maximum speed through the standard SHM formulas.
13.103 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.10
In Exercise 13.9, let us take the position of the mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the 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 in initial phase?
Hint. Use the same omega = 20 rad/s and A = 2 cm from Q13.9 throughout; only the starting point changes which trig function and sign is needed.
Using omega = 20 rad/s and A = 2 cm from Q13.9:

(a) Starting at the mean position, moving in the positive direction: x(t) = 2 sin(20t) cm.

(b) Starting at maximum stretch (positive extreme): x(t) = 2 cos(20t) cm.

(c) Starting at maximum compression (negative extreme): x(t) = -2 cos(20t) cm.

✦ All three functions share the same angular frequency (20 rad/s) and the same amplitude (2 cm) — they differ ONLY in initial phase, since the physical spring-mass system is identical in all three cases and only the moment chosen as t = 0 changes.
13.114 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.11
Two circular motions are shown: (a) radius 3 cm, period 2 s, particle P starting at t=0 at the bottom of the circle (on the -y axis), rotating anticlockwise. (b) radius 2 m, period 4 s, particle P starting at t=0 at the left of the circle (on the -x axis), rotating clockwise. Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Hint. Identify the starting angle from the +x axis and the sense of rotation for each circle, then write x(t) = A cos(omega t + phi), flipping the sign inside for clockwise rotation.
(a) A = 3 cm, omega = 2 pi/T = 2pi/2 = pi rad/s. P starts at the bottom, angle phi = -pi/2 from the +x axis, rotating anticlockwise (the standard direction the formula assumes).

x(t) = 3 cos(pi t - pi/2) = 3 sin(pi t) cm.

(b) A = 2 m, omega = 2pi/T = 2pi/4 = pi/2 rad/s. P starts at the left, angle phi = pi, rotating CLOCKWISE, which reverses the sign inside the cosine compared to the standard anticlockwise case.

x(t) = 2 cos(-(pi/2)t + pi) = -2 cos((pi/2)t) m.

✦ Case (b) needs the extra sign flip inside the cosine because its rotation is clockwise rather than anticlockwise — checking the sense of rotation before writing x(t) is exactly the step that is easy to skip and get the sign wrong.
13.124 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.12
For each of the following simple harmonic motions, find the radius of the reference circle, the initial (t=0) position of the particle on it, and the angular speed of the rotating particle (rotation is 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).
Hint. Convert each function into the standard A cos(omega t + phi) form using -sin(theta) = cos(theta + pi/2) and cos(-theta) = cos(theta), then read off A, omega, and phi directly.
(a) -2 sin(3t + pi/3) = 2 cos(3t + pi/3 + pi/2) = 2 cos(3t + 5pi/6). Radius = 2 cm, angular speed = 3 rad/s, initial angle = 5 pi/6 (150 degrees).

(b) cos(pi/6 - t) = cos(t - pi/6), since cosine is even. Radius = 1 cm, angular speed = 1 rad/s, initial angle = -pi/6 (-30 degrees).

(c) 3 sin(2 pi t + pi/4) = 3 cos(2 pi t + pi/4 - pi/2) = 3 cos(2 pi t - pi/4). Radius = 3 cm, angular speed = 2 pi rad/s, initial angle = -pi/4 (-45 degrees).

(d) 2 cos(pi t) is already in standard form. Radius = 2 cm, angular speed = pi rad/s, initial angle = 0 (starts on the +x axis).

✦ Each part reduces to reading A, omega, and phi off the standard A cos(omega t + phi) form directly, once the given function is rewritten into that exact form using basic trig identities.
13.135 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.13
A spring of force constant k is clamped rigidly at one end with a mass m attached to its free end; a force F applied at the free end stretches the spring. The same spring, with both ends free, is attached to a mass m at either end, and each end is stretched by the same force F. (a) What is the maximum extension of the spring in the two cases? (b) If the mass in the first case and the two masses in the second case are released, what is the period of oscillation in each case?
Hint. For the extension, think about the tension throughout the spring in each setup. For the period of the two-free-masses case, use that the centre of mass between two equal masses does not move, so each mass effectively oscillates about a fixed point using the reduced mass.
(a) Maximum extension: in BOTH cases the spring carries an internal tension of F throughout its length (one end fixed and pulled with F, or both ends pulled outward with F, are physically equivalent for the spring itself), so extension = F/k in both cases.

(b) Case (a), one end fixed: this is the standard single-mass spring system. T_a = 2 pi sqrt(m/k).

Case (b), both ends free with equal mass m at each end: since the two masses are equal, the centre of mass stays fixed, so each mass oscillates about that fixed point exactly as if attached to a spring of constant k anchored there, but with the REDUCED mass mu = m x m/(m+m) = m/2 governing the motion. T_b = 2 pi sqrt(mu/k) = 2 pi sqrt(m/2k).

✦ The extension is identical in both cases since it depends only on the spring's internal tension, but the period is shorter in the two-mass case (by a factor of sqrt(2)) because the reduced mass m/2 that actually governs the oscillation is smaller than the full mass m used in the single-mass case.
13.142 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.14
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
Hint. Stroke is twice the amplitude, so find A first, and keep the angular frequency in the same time unit (per minute) throughout, or convert consistently to per second.
Step 1 — Amplitude. Stroke = 2A = 1.0 m, so A = 0.5 m.

Step 2 — Maximum speed. v_max = omega A = 200 rad/min x 0.5 m = 100 m/min.

Step 3 — Convert to m/s. v_max = 100/60 = 1.67 m/s.

✦ The piston's maximum speed is about 1.67 m/s, found directly from v_max = omega A once the stroke length is correctly halved to get the amplitude.
13.153 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.15
The acceleration due to gravity on the surface of the moon is 1.7 m/s^2. What is the time period of a simple pendulum on the surface of the moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m/s^2.)
Hint. The pendulum's length L is the same on both the Earth and the Moon; only g changes, so take the ratio of the two period formulas to eliminate L entirely.
Step 1 — Ratio of periods. T = 2 pi sqrt(L/g), so T_moon/T_earth = sqrt(g_earth/g_moon).

Step 2 — Substitute values. T_moon = 3.5 x sqrt(9.8/1.7) = 3.5 x sqrt(5.7647) = 3.5 x 2.401 = 8.40 s.

✦ The same pendulum swings with a period of about 8.4 s on the Moon, more than twice as long as on Earth, since a smaller g means a weaker restoring force and therefore a slower oscillation for the same length.
13.164 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.16
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?
Hint. In the car's frame, the bob experiences both real gravity (downward) and a centrifugal pseudo-force (horizontal, magnitude v-squared over R) — combine these into a single effective gravity before applying the pendulum formula.
Step 1 — Identify the effective gravity. In the car's rotating frame, the bob experiences real gravity g downward and a centrifugal pseudo-force per unit mass of v^2/R directed horizontally outward from the track's centre. These combine as perpendicular vector components.

Step 2 — Combine them. g_eff = sqrt(g^2 + (v^2/R)^2).

Step 3 — Apply the pendulum formula with g_eff. T = 2 pi sqrt(l/g_eff) = 2 pi sqrt(l / sqrt(g^2 + v^4/R^2)).

✦ The period is shorter than a stationary pendulum's, since g_eff is always larger than g alone — the centrifugal pseudo-force adds to the restoring effect available to the pendulum, strengthening the effective gravity it feels.
13.175 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.17
A cylindrical piece of cork of base area A, height h, and density (of the cork) rho 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(h rho / (rho_l g)). (Ignore damping due to viscosity of the liquid.)
Hint. Find the equilibrium submerged depth first from the weight-equals-buoyancy condition, then find the NET restoring force for a small additional displacement y below equilibrium.
Step 1 — Equilibrium condition. At equilibrium, weight equals buoyant force: (A h rho) g = A x0 rho_l g, where x0 is the equilibrium submerged depth. This just fixes x0; it is not needed further.

Step 2 — Restoring force for a small extra displacement y. If the cork is pushed down an additional distance y, the EXTRA buoyant force (beyond equilibrium) is A y rho_l g, acting upward — a net restoring force.

F = -A rho_l g y

Step 3 — Apply Newton's second law. The cork's total mass is A h rho (constant, since h and rho do not change).

(A h rho) a = -A rho_l g y, so a = -(rho_l g / (h rho)) y

This is the SHM form a = -omega^2 y, with omega^2 = rho_l g / (h rho).

Step 4 — Period. T = 2 pi / omega = 2 pi sqrt(h rho / (rho_l g)).

✦ This matches the required expression exactly, since the restoring force comes entirely from the CHANGE in buoyant force due to the extra submerged depth y, while the cork's own weight (and its full mass A h rho) stays fixed throughout the motion.
13.185 marksNCERT Cl-11 Physics Part II, Ch13 Exercises, Q13.18
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.
Hint. Let the mercury level on one side rise by y above equilibrium; by conservation of volume in a uniform tube, the other side must fall by the same y, making the height DIFFERENCE between the two arms equal to 2y — this height difference is what drives the restoring force.
Step 1 — Set up the displacement. Let the mercury be displaced so one arm's level is y above the equal-level equilibrium and the other is y below it. The height difference between the two arms is 2y.

Step 2 — Restoring force from the height difference. The unbalanced pressure due to this height difference is rho g (2y), where rho is mercury's density. Acting over the tube's cross-sectional area A, the net restoring force on the whole mercury column is:

F = -2 rho g A y

Step 3 — Apply Newton's second law to the full mercury column. Let L be the total length of mercury in the tube (constant, since mercury is incompressible and the tube's cross-section is uniform). Total mass = rho A L.

(rho A L) a = -2 rho g A y, so a = -(2g/L) y

This is the SHM form a = -omega^2 y, with omega^2 = 2g/L.

Step 4 — Conclusion. T = 2 pi sqrt(L/2g), confirming SHM with a well-defined period.

✦ The mercury oscillates because any height imbalance between the two arms creates a restoring pressure difference proportional to that imbalance, which is exactly the linear-restoring-force condition that defines SHM.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part II textbook, Reprint 2026-27 (keph206.pdf, 19 pages, "CHAPTER THIRTEEN"), cross-checked against the official CBSE curriculum 2026-27, Subject Code 042, Unit X. One end-of-chapter Exercises set (18 questions, 13.1-13.18). No content gap found against CBSE's syllabus line. One textual leftover noted: the chapter's introduction promises damped and forced oscillations 'later in the chapter', but the rationalised edition's section list stops at 13.8 and neither topic appears in the body or in CBSE's syllabus line for this chapter.. Questions are referenced from the NCERT textbook for identification.

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