A particle with a mass of$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{20}}\mathbf{}\mathbf{kg}$is oscillating with simple harmonic motion with a period of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{s}$and a maximum speed of$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$.

1. Calculate the angular frequency of the particle.
2. Calculate the maximum displacement of the particle.

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

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

$\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}$ (i)

The velocity of a body undergoing simple harmonic motion

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

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

$\begin{array}{rcl}\mathrm{\omega }& =& \frac{2\mathrm{\pi }}{1\times {10}^{-5}}\\ & =& 6.28\times {10}^5\mathrm{rad}/\sec \end{array}$

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

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

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

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