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?

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

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)

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

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