Question: A block of mass 200 g executing SHM under the influence of a spring of spring constant k = 90 N m–1 ...

Question

A block of mass 200 g executing SHM under the influence of a spring of spring constant k = 90 N m–1 and a damping constant b = 40 g s–1. The time elapsed for its amplitude to drop to half of its initial value is (Given ln(1/2) = –0.693)

Answer 1

Solution

The amplitude of the damped oscillator at any instant t is given by

Where is its initial amplitude and b is the damping constant.

At ,the amplitude drop to half of its initial

value from (i), we get

Taking natural logarithm on both sides, we get

In $\left( \frac { 1 } { 2 } \right) = \frac { \mathrm { bt } _ { 1 / 2 } } { 2 \mathrm {~m} }$

$t _ { 1 / 2 } = - \frac { 2 m \ln ( 1 / 2 ) } { b }$ ….. (ii)

Here, in

Substituting these values in eq. (ii) we get