Question

A block of mass $200$ g executing SHM under the influence of a spring of spring constant $k=90N{m}^{-1}$ and a damping constant $b=40g{s}^{-1}$. Time taken for its amplitude of vibrations to drop to half of its initial values (Given, In $\left(1/2\right)=-0.693)$

• A. $7$s
• B. $9$s
• C. $4$s
• D. $11$s

Answer 1

The correct option is A $7$s

Given data,

mass $m=200g$

Spring constant $k=90N{m}^{-1}$

Damping constant $b=40g{s}^{-1}$

To calculate: Time taken for the amplitude of vibration to drop to half of the initial value

We know that amplitude at any time t can be given as:

$A\left(t\right)=A_0{e}^{-\frac{bt}{2m}}$

or $T_{1/2}=\frac{-0.693l\times 2\times 0.2}{40\times {10}^{-3}}=6.93s$

Time taken for its amplitude of vibrations to drop to half of its initial values is $7s$