91Ó°ÊÓ

The amplitude of the diaphragm,${\mathrm{X}}_{\mathrm{m}}=1\mathrm{μ³¾}\mathrm{or}{10}^{-6}\mathrm{m}$

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,

${\mathrm{a}}_{\mathrm{m}}={\mathrm{Ó\neg }}^2{\mathrm{X}}_{\mathrm{m}}$ …(¾±)

Angular frequency of a body in oscillation,

$\mathrm{Ó\neg }=2\mathrm{Ï„}\mathrm{Ï„}f$ …(¾±¾±)

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

$a={\mathrm{Ó\neg }}^2\left(1×{10}^{-6}\right)$

$$\begin{gathered} 9.8\mathrm{m}/{\mathrm{s}}^2={\mathrm{Ó\neg }}^2\left(1×{10}^{-6}\mathrm{m}\right)\left(âˆ\mu \mathrm{given},\mathrm{a}=\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{Ó\neg }=3130.495\mathrm{rad}/\sec \end{gathered}$$

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

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

Hence, the frequency value is found to be 498.5 Hz

From equation (i)

$a\mathrm{Î\pm }{\mathrm{Ó\neg }}^2$

And $\mathrm{Ó\neg }\mathrm{Î\pm }f$

So that,

$a\mathrm{Î\pm }{f}^2$

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