Question

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

(a) At what frequency is the magnitude a of the diaphragm's acceleration equal to g?

Answer 1

Frequency will be equal to 361.64 Hz

Step-by-step explanation:

We have given amplitude A =
`1.9\mu m=1.9* 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* 1.9* 10^(-6)`

`\omega ^2=51.57* 10^6`

`\omega =2.271* 10^3rad/sec`

Now frequency
`f=(\omega )/(2\pi )=(2271)/(2* 3.14)=361.64Hz`