Question

the amplitude of an oscillator decreases to 36.8% of its initial value in 10.0 s. what is the value of the time constant

Answer 1

Answer:

τ = 5 s

Explanation:

When a vibrating body is damped. Its amplitude starts to decrease. This decrement is exponential. And it is given as follows:

$X = X_{0}e^{-\frac{t}{2\tau}$

where,

τ = Time Constant = ?

X = Instantaneous value of amplitude

X₀ = Initial Value of amplitude

t = time interval = 10 s

The ratio of decrement is given as:

$\frac{X}{X_0} = 36.8\% = 0.368$

therefore, using these values, we get:

$\frac{X}{X_{0}} = 0.368 = e^{\frac{10\ s}{2\tau}}$

Taking natural log (ln) on both sides, we get:

$ln(0.368) = \frac{10\ s}{2\tau}\\\\\tau = \frac{10\ s}{2ln(0.368)}$

τ = 5 s