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Expert-verified Found in: Page 436 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 120 Hz.Find the amplitude.Find the maximum blade speed. Find the magnitude of the maximum blade acceleration.

1. Amplitude is $1\mathrm{mm}\mathrm{or}{10}^{-3}\mathrm{m}$
2. Maximum blade speed is $0.75\mathrm{m}/\mathrm{s}$
3. Magnitude of maximum blade acceleration is $567.461\mathrm{m}/{\mathrm{s}}^{2}$.
See the step by step solution

## Step 1: Stating the given data

1. Back and forth distance, $\mathrm{d}=2\mathrm{mm}$
2. Frequency of the body, $\mathrm{f}=120\mathrm{Hz}$.

## Step 2: Understanding the concept of motion

The amplitude is half of the back-and-forth distance. The velocity is maximum when the displacement is zero. We can use this concept to find the maximum velocity. The acceleration is maximum when the displacement is maximum, and we can use this concept to find the maximum acceleration.

Formulae:

Angular frequency of a body in oscillation

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

The velocity of the body in motion

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

Acceleration of body in simple harmonic motion

${\mathrm{a}}_{\mathrm{m}}={\mathrm{\omega }}^{2}{\mathrm{X}}_{\mathrm{m}}$ (iii)

## Step 3: a) Calculation of amplitude

The amplitude of oscillation is half of the back-and-forth distance.

$\begin{array}{rcl}{\mathrm{X}}_{\mathrm{m}}& =& \mathrm{d}/2\\ & =& 1\mathrm{mm}\mathrm{or}{10}^{-3}\mathrm{m}\end{array}$

Hence, the amplitude of the body is $1\mathrm{mm}\mathrm{or}{10}^{-3}\mathrm{m}$.

## Step 4: b) Calculation of maximum blade velocity

From equation (i), we get the angular frequency as

$\begin{array}{rcl}\mathrm{\omega }& =& 2\mathrm{\pi }×120\\ & =& 753.3\mathrm{rad}/\mathrm{sec}\end{array}$

So, the blade velocity, using equation (ii) and the given values, is given as follows:

$\begin{array}{rcl}\mathrm{v}& =& 753.3×0.001\\ & =& 0.75\mathrm{m}/\mathrm{s}\end{array}$

Hence, the blade velocity is $0.75\mathrm{m}/\mathrm{s}$.

## Step 5: c) Calculation of magnitude of maximum blade acceleration

Using equation (iii) and the given values, we get the acceleration of a body as

$\begin{array}{rcl}{\mathrm{a}}_{\mathrm{m}}& =& 753.{3}^{2}×0.001\\ & =& 567.461\mathrm{m}/{\mathrm{s}}^{2}\end{array}$

In two significant figures

${\mathrm{a}}_{\mathrm{m}}=5.7×{10}^{2}\mathrm{m}/{\mathrm{s}}^{2}$

Hence, the value of acceleration of the body is $567.461\mathrm{m}/{\mathrm{s}}^{2}$.

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