Answer
0.25 s
Work Step by Step
Here we use the equation, $x(t)=Ae^{\frac{-bt}{2m}}cos(\omega t+\theta)$; Where, x(t) - Displacement, A - Amplitude, $\omega $ -Angular velocity, t - time, b - constant, m - mass on the string, $\theta$ - Phase constant.
$x(t)=Ae^{\frac{-bt}{2m}}cos(\omega t+\theta)-(1)$
When the amplitude drops to half its initial value, we can write,
$e^{\frac{-bt}{2m}}=\frac{1}{2}$
Now Let's take the natural logarithm of both sides.
$lne^{\frac{-bt}{2m}}=ln\frac{1}{2}=\gt \frac{-bt}{2m}=-0.7-(2)$
We know that the $\frac{b}{2m}=2.8\space s^{-1}=\gt(2)$
$2.8\space s^{-1}t=0.7$
$t=0.25\space s$
It will take 0.25 s amplitude to drop to half its original value.
