Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to drop to half of its initial value? (ln 2 = 0.693)

Solution

Physics sir · 4 steps

Amplitude decays as e^(-bt/2m), so set that factor to one half and solve for t.

1. Damped amplitude. With a damping force -bv, the amplitude falls exponentially with time. The exponent has 2m in the denominator, not m.

   A = A₀ e^(-bt/2m)

2. Half amplitude. Put A = A₀/2. Then e^(-bt/2m) = 1/2, so bt/2m = ln 2 and t = (2m ln 2)/b.
3. Put in numbers. Keep both quantities in grams: m = 500 g and b = 20 g s⁻¹. Then t = (2 × 500 × 0.693)/20 = 693/20 = 34.65 s.
4. So. The amplitude drops to half its initial value after 34.65 s, which is option (b).

The idea to keep. For damped SHM, amplitude goes as e^(-bt/2m) and energy as e^(-bt/m). Half-amplitude time is (2m ln 2)/b. Use the same mass unit in m and b.

Answer from the NTA final answer key. Solution by PrepWiser.