Oscillations — Practice

Question 1

A particle's acceleration and displacement are related by a = -5x (SI units). Is this SHM? If so, find the angular frequency.

Accepted Answer

Yes — a is linear in x with a negative constant, matching a = -omega^2 x. omega^2 = 5, so omega = sqrt(5) rad/s.

Question 2

A particle completes 20 oscillations in 4 seconds. Find its frequency and period.

Accepted Answer

nu = 20/4 = 5 Hz. T = 1/nu = 1/5 = 0.2 s.

Question 3

Classify each as periodic or non-periodic: (a) sin(omega t), (b) e^(-t).

Accepted Answer

(a) Periodic (in fact SHM) — repeats every 2 pi/omega. (b) Non-periodic — it decays monotonically toward zero and never repeats a value.

Question 4

A particle executes SHM with angular frequency omega = 2 rad/s. At t = 0, its displacement is 3 cm and its velocity is -8 cm/s. Find the amplitude and phase constant using x(t) = A cos(omega t + phi).

Accepted Answer

A cos(phi) = 3 and -omega A sin(phi) = -8, giving A sin(phi) = 4. A = sqrt(3^2+4^2) = 5 cm. tan(phi) = 4/3, and since both sin and cos of phi are positive, phi = arctan(4/3) is approximately 53.1 degrees (0.927 rad).

Question 5

A particle executes SHM: x(t) = 4 cos(2t + pi/6) cm. Find its velocity and acceleration at t = pi/6 s.

Accepted Answer

At t = pi/6 s, the phase is 2(pi/6) + pi/6 = pi/2. x = 4cos(pi/2) = 0 (mean position). v = -omega A sin(pi/2) = -2 x 4 x 1 = -8 cm/s. a = -omega^2 x = -4 x 0 = 0.

Question 6

A 0.5 kg mass attached to a spring oscillates with a period of 0.2 pi s. Find the spring constant.

Accepted Answer

T = 2 pi sqrt(m/k), so k = m(2 pi/T)^2 = 0.5 x (2 pi / 0.2 pi)^2 = 0.5 x 10^2 = 50 N/m.

Question 7

A particle of mass 2 kg executing SHM has total energy 16 J and amplitude 0.4 m. Find the force constant k, and the kinetic energy when the displacement is 0.2 m.

Accepted Answer

E = (1/2)kA^2, so k = 2 x 16 / 0.16 = 200 N/m. At x = 0.2 m: U = (1/2)(200)(0.04) = 4 J. K = E - U = 16 - 4 = 12 J.

Question 8

Find the length of a simple pendulum whose period is exactly 4 s on Earth's surface (g = 9.8 m/s^2).

Accepted Answer

L = g T^2 / (4 pi^2) = 9.8 x 16 / 39.48 = 3.97 m.

Question 9

A simple pendulum of length 1.0 m is taken to the Moon, where g = 1.7 m/s^2. (a) Find its period on the Moon. (b) By what factor does the period change compared to Earth (g = 9.8 m/s^2)?

Accepted Answer

(a) T = 2 pi sqrt(1/1.7) = 2 pi x 0.767 = 4.82 s. (b) Factor = sqrt(g_earth/g_moon) = sqrt(9.8/1.7) = 2.40 — the period on the Moon is about 2.4 times longer.

Question 10

Two identical springs, each of spring constant k, are connected in series and attached to a mass m at the free end, with the other end of the combination fixed to a wall. Find the period of oscillation.

Accepted Answer

For springs in series, 1/k_eff = 1/k + 1/k = 2/k, so k_eff = k/2. T = 2 pi sqrt(m/k_eff) = 2 pi sqrt(2m/k).

Practice — Oscillations

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