In an electric shaver, the blade moves back and forth over a distance of 2.0 mmin simple harmonic motion, with frequency 120 Hz.

1. Find the amplitude.
2. Find the maximum blade speed.
3. Find the magnitude of the maximum blade acceleration.

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

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)

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 $1\mathrm{mm}\mathrm{or}{10}^{-3}\mathrm{m}$.

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

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

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

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

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

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

$\begin{array}{rcl}{\mathrm{a}}_{\mathrm{m}}& =& 753.3^2\times 0.001\\ & =& 567.461\mathrm{m}/{\mathrm{s}}^2\end{array}$

In two significant figures

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

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