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Found in: Page 436

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

1. The angular frequency of the particle is $6.28×{10}^{5}\mathrm{rad}/\mathrm{sec}$.
2. The maximum displacement of the particle is $1.59×{10}^{-3}\mathrm{m}$.
See the step by step solution

## Step 1: Stating the given data

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

## Step 2: Understanding the concept of motion

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)

## Step 3: a) Calculation of angular frequency of the body

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

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

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

## Step 4: b) Calculation of maximum displacement

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

$1×{10}^{3}=6.28×{10}^{5}×{\mathrm{X}}_{\mathrm{m}}\phantom{\rule{0ex}{0ex}}{\mathrm{X}}_{\mathrm{m}}=1.59×{10}^{-3}\mathrm{m}$

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

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