Question

A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to 1.9 μm.

Answer 1

Answer:

Frequency will be equal to 361.64 Hz

Explanation:

We have given amplitude A = $1.9\mu m=1.9\times 10^{-6}m$

Acceleration is equal to g

Value of $g=9.8m/sec^2$

Acceleration is equal to $a=\omega ^2A$

$9.8=\omega ^2\times 1.9\times 10^{-6}$

$\omega ^2=51.57\times 10^6$

$\omega =2.271\times 10^3rad/sec$

Now frequency $f=\frac{\omega }{2\pi }=\frac{2271}{2\times 3.14}=361.64Hz$