91Ó°ÊÓ

1. Mass of particle,$\mathrm{m}=1×{10}^{-20}\mathrm{kg}$
2. Time period of SHM,data-custom-editor="chemistry" $\mathrm{t}=1×{10}^{-5}\sec$
3. Maximum speed of the body,$\mathrm{v}=1×{10}^3\mathrm{m}/\mathrm{s}$

We can find the angular frequency by using the time period of motion. And maximum displacement can be found by using the formulae of maximum velocity and angular frequency.

Formula:

Angular frequency of a body in oscillation

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{\mathrm{T}}$ (i)

The velocity of a body undergoing simple harmonic motion

$\mathrm{v}={\mathrm{Ó\neg X}}_{\mathrm{m}}$ (ii)

Using equation (i) and the given values, the angular frequency is given as follows:

$\begin{array}{rcl}\mathrm{Ó\neg }& =& \frac{2\mathrm{Ï€}}{1×{10}^{-5}}\\ & =& 6.28×{10}^5\mathrm{rad}/\sec \end{array}$

Hence, the angular frequency is $6.28×{10}^5\mathrm{rad}/\sec$.

Using the given values and equation (ii), we get the maximum displacement as follows:

$$\begin{gathered} 1×{10}^3=6.28×{10}^5×{\mathrm{X}}_{\mathrm{m}} \\ {\mathrm{X}}_{\mathrm{m}}=1.59×{10}^{-3}\mathrm{m} \end{gathered}$$

Hence, the maximum displacement is $1.59×{10}^{-3}\mathrm{m}$.