Previous year questions: Oscillations

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar for this unit. Each one uses the formulas on the first page of this chapter, and every answer is worked out in full so that the method behind the correct option is clear.

Q1 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

The velocity of a particle in SHM along the x-axis obeys v squared = 50 - x squared. If the period of the motion is x/7 seconds, find x.

Answer: Comparing v squared = omega squared (A squared - x squared) with v squared = 50 - x squared gives omega squared = 1, so omega = 1 rad s^-1. The period is 2 pi/omega = 2 pi s. Setting x/7 equal to 2 pi gives x = 14 pi = 44 (to the nearest integer). The option is 44. Note that the given equation already contains the angular frequency, so the amplitude follows as sqrt 50 = 7.07.

Q2 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

A uniform disc of radius R and mass M oscillates about an axis through a point A on its rim, the axis lying in the plane of the disc. Find the period of small oscillations.

Answer: This is a physical pendulum about an axis at distance R from the centre. The moment of inertia about the diameter through A is (1/4)M R squared, and by the parallel-axis theorem about the axis at the rim the value is (1/4)M R squared + M R squared = (5/4)M R squared. The period is 2 pi sqrt(I/(M g R)) = 2 pi sqrt((5R/4g)). The correct option is 2 pi sqrt(5R/4g). The result is larger than for a simple pendulum of length R because the disc's moment of inertia is bigger than that of a point mass at the rim.

Q3 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Match List I with List II: A. sin squared(omega t), B. sin cubed(2 omega t), C. sin(omega t) + cos(pi omega t), D. cos(omega t) + cos(2 omega t) against I. periodic with period pi/omega but not SHM, II. periodic with period 2 pi/omega but not SHM, III. periodic with period pi/omega and SHM, IV. non-periodic.

Answer: sin squared(omega t) = (1 - cos 2 omega t)/2 is sinusoidal about its mean value, so it is SHM with period pi/omega: A matches III. sin cubed(2 omega t) is a sum of sin 2 omega t and sin 6 omega t terms, periodic with period pi/omega but not SHM: B matches I. sin(omega t) + cos(pi omega t) has two incommensurate frequencies, so it is non-periodic: C matches IV. cos(omega t) + cos(2 omega t) is periodic with period 2 pi/omega but is not SHM because it has two frequencies: D matches II. The correct option is A-III, B-I, C-IV, D-II. The key test for SHM is whether the motion can be written as a single sine or cosine of time.

Q4 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

A particle executes SHM with amplitude A and time period 5 s. Find the time it takes to move from x = A to x = A/sqrt(2).

Answer: Using x = A cos(2 pi t/T), the position x = A/sqrt(2) gives cos(2 pi t/5) = 1/sqrt(2), so 2 pi t/5 = pi/4 and t = 5/8 s = 0.625 s. The particle moves from the extreme through a phase of pi/4, which is one eighth of the full cycle, and one eighth of 5 s is 5/8 s. The correct answer is 5/8 s. If the particle started from the mean position instead, the same interval would take 1/8 s, so the starting point decides the phase.

Q5 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

The frequency of oscillation of a mass m suspended by a spring is v1. If the spring is cut in half and the same mass is suspended from one half, the frequency becomes v2. Find v2/v1.

Answer: The spring constant of a spring is inversely proportional to its natural length, so cutting it in half doubles the spring constant: k2 = 2k1. The frequency is f = (1/2 pi) sqrt(k/m), so v2/v1 = sqrt(k2/k1) = sqrt(2). The correct option is sqrt 2. A spring is made of many coils in series, and each half is stiffer than the whole.

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

A rigid dipole executes simple harmonic motion about its centre in an electric field E1 = E0 x-hat. A second field E2 = 2E0( y-hat + z-hat ) is switched on. Find the approximate percentage change in the frequency of oscillation.

Answer: For a dipole in an electric field the restoring torque is proportional to the field magnitude, so the frequency is proportional to the square root of the effective field. The resultant field has magnitude E0 sqrt(1 squared + 2 squared + 2 squared) = 3E0. The new frequency is sqrt(3) times the old one, so the percentage increase is (sqrt 3 - 1) x 100 = 73 percent. The correct option is 73 percent. The frequency scales as the square root of the restoring constant, so a field three times as strong gives a frequency sqrt 3 times as large.

Q7 (NCERT Exemplar, 2 marks)

The displacement of a particle is given by y = sin cubed(omega t). Describe the motion and give its period.

Answer: Use the identity sin cubed theta = (3 sin theta - sin 3 theta)/4, so y is a sum of sin(omega t) and sin(3 omega t). Each term is periodic with a period that divides 2 pi/omega, and the fundamental period is 2 pi/omega. The motion is periodic but not simple harmonic, because it is a sum of two harmonic terms and not a single sine or cosine. The correct choice is that the motion is periodic but not simple harmonic. The fundamental period is the time in which both terms complete a whole number of cycles.

Q8 (NCERT Exemplar, 1 mark)

Which of the following relations between acceleration and displacement represents simple harmonic motion: a = +2x, a = +2x squared, a = -2x squared, or a = -2x?

Answer: SHM requires the acceleration to be proportional to the displacement and opposite in sign, that is a = -constant times x. Only a = -2x has this form, with the constant 2 equal to omega squared. The choices with x squared are not linear in the displacement, and a = +2x gives repulsive rather than restoring behaviour. The correct choice is a = -2x. The constant 2 is the square of the angular frequency, so the period is 2 pi/sqrt 2.

Q9 (NCERT Exemplar, 1 mark)

The motion of an oscillating liquid column in a U-tube is best described as which of the following?

Answer: The weight of the excess liquid in one arm provides a restoring force proportional to the displacement of the column, so the motion is simple harmonic. The period contains g and the length of the column, but the density of the liquid cancels because both the restoring force and the mass are proportional to it. The correct choice is that the motion is simple harmonic and the period is independent of the density of the liquid. Only the geometry of the column and the value of g fix the period.

Q10 (NCERT Exemplar, 2 marks)

The displacement of a particle is y = a sin(omega t) + b cos(omega t). What is the nature of the motion and what is its amplitude?

Answer: Write the sum as a single cosine: a sin(theta) + b cos(theta) = R cos(theta - alpha) with R = sqrt(a squared + b squared). The displacement is therefore a single sinusoidal function of time, so the motion is simple harmonic with the same angular frequency omega and with amplitude sqrts of a squared plus b squared. The correct choice is SHM with amplitude sqrt(a squared + b squared). The phase constant is the angle whose tangent is b/a. The two amplitudes add in quadrature rather than directly, because the sine and cosine parts are a quarter cycle apart.