Question Detail

PHY 104 PHY 104 Past Questions Objective Question

Question

The Q-factor of an underdamped SHO of frequency 600 Hz is 9000. What number of oscillations will it perform for its amplitude to decay to 1/e of its initial value?
Options
A 2547
Correct Answer
B 2685
C 2865
D 2950
Correct Answer

Option A is the correct answer.

Detailed Explanation

The Q-factor is defined as Q = 2π × (energy stored)/(energy dissipated per cycle). It is also related to the decay of amplitude: A(t) = A₀ e^(-γt/2), where γ is the damping coefficient. The number of oscillations for amplitude to decay to 1/e is Q/π. So N = Q/π = 9000/π ≈ 2864.8. The closest option is 2865. Let's check: Q = 2πf/γ. The amplitude decays as e^(-γt/2). For it to decay to 1/e, γt/2 = 1 => t = 2/γ. Number of oscillations N = f t = f(2/γ) = (2f)/γ = Q/π. So N = 9000/π = 2864.8 ≈ 2865.

Hint

For an underdamped oscillator, the number of oscillations for amplitude to decay to 1/e is Q/π.

Question Info