Question

A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu m}$.

(a) At what frequency is the magnitudeof the diaphragm’s acceleration equal to g?

(b) For greater frequencies, isgreater than or less than g?

Short answer

a) The frequency at which the magnitude of the acceleration is equal to g is 498.8 Hz

b) For greater frequencies a will be greater than g.

Step-by-step solution

The given data

The amplitude of the diaphragm,${\mathrm{X}}_{\mathrm{m}}=1\mathrm{\mu m}\mathrm{or}{10}^{-6}\mathrm{m}$

Understanding the concept of motion

Maximum acceleration is a product of the square of angular frequency and amplitude. We can use this concept to find the frequency of a body in simple harmonic motion.

Formula:

Acceleration of body undergoing simple harmonic motion,

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

Angular frequency of a body in oscillation,

$\omega =2\tau \tau f$ …(ii)

(a) Calculation of frequency of body in oscillation

Using equation (i) and the given values, we get the angular frequency of body as:

$a={\omega }^2\left(1\times {10}^{-6}\right)$

$$\begin{gathered} 9.8\mathrm{m}/{\mathrm{s}}^2={\mathrm{\omega }}^2\left(1\times {10}^{-6}\mathrm{m}\right)\left(\because \mathrm{given},\mathrm{a}=\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{\omega }=3130.495\mathrm{rad}/\sec \end{gathered}$$

Using equation (ii), the frequency of body is given as:

$$\begin{gathered} \mathrm{f}=\frac{3132.495\mathrm{rad}/\sec }{2\mathrm{\tau \tau }} \\ =498.5\mathrm{Hz} \end{gathered}$$

Hence, the frequency value is found to be 498.5 Hz

(b) Studying the effect of increase in frequency on acceleration

From equation (i)

$a\alpha {\omega }^2$

And $\omega \alpha f$

So that,

$a\alpha f^2$

As acceleration is directly proportional to square of frequency, if we increase frequency then acceleration also increases and it will be greater than g