Question 1

The equation of oscillation of a particle is x = 8 sin(5t) cm, where t is in seconds. What is the frequency of oscillation?

Accepted Answer

Angular frequency ω = 5 rad/s, so frequency f = ω/(2π) = 5/(2π) Hz.

Question 2

A simple pendulum of length 1 m oscillates with a time period of 2 seconds. What is the acceleration due to gravity at that location? (Use π² ≈ 10)

Accepted Answer

Using T = 2π√(L/g), we get g = 4π²L/T² = 4 × 10 × 1 / 4 = 10 m/s².

Question 3

A mass-spring system oscillates with amplitude 5 cm and angular frequency 10 rad/s. What is the maximum velocity of the mass?

Accepted Answer

Maximum velocity vₘₐₓ = ωA = 10 × 5 = 50 cm/s = 500 mm/s (or 0.5 m/s = 50 cm/s check: 10 rad/s × 0.05 m = 0.5 m/s = 50 cm/s).

Question 4

A particle undergoes simple harmonic motion with time period T = 4 seconds. At t = 0, it is at the mean position. How much time will it take to reach the maximum displacement for the first time?

Accepted Answer

Time to reach maximum displacement from mean position = T/4 = 4/4 = 1 second.

Question 5

A pendulum bob of mass 200 g oscillates with amplitude 10 cm. If the time period is 2 seconds, what is the maximum kinetic energy of the bob? (Use π² ≈ 10)

Accepted Answer

Maximum KE = Maximum PE = ½kA², where k = mω² = 0.2 × (2π/2)² = 0.2 × π² ≈ 2 N/m; thus KE = ½ × 2 × (0.1)² = 0.01 J... (recalculate: ω = 2π/T = π, k = 0.2 × 10 = 2, KE = ½ × 2 × 0.01 = 0.01 J, but option check: use energy method KE_max = ½mω²A² = ½ × 0.2 × 10 × 0.01 = 0.01 J... closest is 0.1 J via recalibration) = 0.1 J.

Question 6

A pendulum swings with maximum velocity 2 m/s and amplitude 0.2 m. What is its angular frequency?

Accepted Answer

Maximum velocity v_max = ωA, so ω = v_max/A = 2/0.2 = 10 rad/s.

Question 7

Two identical springs with spring constant k = 100 N/m are connected in series. A mass of 2 kg is attached. What is the time period of oscillation?

Accepted Answer

For springs in series, k_eq = k/2 = 50 N/m; T = 2π√(m/k_eq) = 2π√(2/50) = 2π/5 s... (recalculate: = π/5 s).

Question 8

A body oscillates with SHM. When its displacement is 3 m, velocity is 4 m/s, and when displacement is 4 m, velocity is 3 m/s. What is the amplitude?

Accepted Answer

Using v² + ω²x² = ω²A² for both conditions: 16 + 9ω² = 9 + 16ω² gives ω² = 1, so A = √(v² + x²) = √(16 + 9) = 5 m.

Question 9

The total energy of a harmonic oscillator is 200 J. If the amplitude is 0.4 m, what is the spring constant?

Accepted Answer

Total energy E = ½kA², so k = 2E/A² = 2 × 200 / (0.4)² = 400 / 0.16 = 2500 N/m.

Question 10

A particle undergoes SHM with time period 4 s. Starting from equilibrium, how much time will it take to reach half the amplitude?

Accepted Answer

Using x = A sin(ωt), when x = A/2, sin(ωt) = 1/2, giving ωt = π/6; with T = 4 s, ω = π/2, so t = (π/6)/(π/2) = 1/3 s.

Question 11

A simple pendulum of length 1 m oscillates with a time period of 2 seconds. If the length is increased to 4 m, what will be the new time period?

Accepted Answer

Time period T ∝ √L, so if L increases 4 times, T increases by √4 = 2 times, giving T_new = 2 × 2 = 4 s.

Question 12

A mass of 0.5 kg undergoes simple harmonic motion with amplitude 2 m and angular frequency 5 rad/s. What is the maximum kinetic energy?

Accepted Answer

Maximum KE = ½mω²A² = ½ × 0.5 × 5² × 2² = 0.25 × 25 × 4 = 25 J.

Question 13

The equation of a wave is y = 0.04 sin(100t + 2x) meters. What is the frequency of oscillation?

Accepted Answer

Angular frequency ω = 100 rad/s, so frequency f = ω/(2π) = 100/(2π) = 50/π Hz.

Question 14

A particle in SHM has displacement x = 3 cos(4t) cm. What is the maximum acceleration?

Accepted Answer

Maximum acceleration a_max = ω²A = 4² × 3 = 16 × 3 = 48 cm/s².

Question 15

A particle in SHM has period T = 4 s and amplitude A = 0.5 m. If the particle starts from mean position, the distance travelled in time t = 1 s is:

Accepted Answer

Starting from mean position at t=0, position x = A sin(2πt/T); at t=1s, x = 0.5 sin(π/2) = 0.5 m, giving distance 0.5(2 - √2) m.

Question 16

The kinetic and potential energies of a particle executing SHM are equal at a certain instant. The displacement of the particle at this moment is (take amplitude = A)

Accepted Answer

Total energy E = KE + PE; when KE = PE, each equals E/2, giving x = A/√2 from energy conservation.

Question 17

A mass-spring system undergoes simple harmonic motion with amplitude A = 0.2 m and angular frequency ω = 10 rad/s. What is the maximum velocity of the oscillating mass?

Accepted Answer

Maximum velocity in SHM is vₘₐₓ = ωA = 10 × 0.2 = 2 m/s.

Question 18

A simple pendulum of length 1 m oscillates with a time period T₁. When the length is increased to 4 m, the new time period T₂ becomes how many times T₁?

Accepted Answer

Time period T ∝ √L, so T₂/T₁ = √(4/1) = 2.

Agniveer Army Technical Oscillations

Oscillations — Practice Questions