The amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500 g, Decay constant = 20 g s$$^{-1}$$ then how much time is required for the amplitude of the system to drop to half of its initial value? ($$\ln 2 = 0.693$$)

In a damped mass-spring system, the amplitude decreases exponentially as $$A(t) = A_0\, e^{-bt/(2m)}$$, where $$b$$ is the decay constant and $$m$$ is the mass.

We need the time when the amplitude drops to half its initial value, i.e., $$A(t) = \frac{A_0}{2}$$. Setting up the equation: $$\frac{A_0}{2} = A_0\, e^{-bt/(2m)}$$, which gives $$e^{-bt/(2m)} = \frac{1}{2}$$.

Taking the natural logarithm: $$\frac{bt}{2m} = \ln 2 = 0.693$$.

Solving for $$t$$: $$t = \frac{2m \ln 2}{b} = \frac{2 \times 500 \times 0.693}{20} = \frac{693}{20} = 34.65$$ s.

The time required for the amplitude to drop to half its initial value is $$34.65$$ s.