91影视

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

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,

${\mathrm{a}}_{\mathrm{m}}={\mathrm{蝇}}^2{\mathrm{X}}_{\mathrm{m}}$ 鈥�(颈)

Angular frequency of a body in oscillation,

$蝇=2蟿蟿f$ 鈥�(颈颈)

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

$a={蝇}^2\left(1脳{10}^{-6}\right)$

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

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

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

Hence, the frequency value is found to be 498.5 Hz

From equation (i)

$a伪{蝇}^2$

And $蝇伪f$

So that,

$a伪f^2$

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