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?

Respuesta :

Answer:

Frequency will be equal to 361.64 Hz

Explanation:

We have given amplitude A = [tex]1.9\mu m=1.9\times 10^{-6}m[/tex]

Acceleration is equal to g

Value of [tex]g=9.8m/sec^2[/tex]

Acceleration is equal to [tex]a=\omega ^2A[/tex]

[tex]9.8=\omega ^2\times 1.9\times 10^{-6}[/tex]

[tex]\omega ^2=51.57\times 10^6[/tex]

[tex]\omega =2.271\times 10^3rad/sec[/tex]

Now frequency [tex]f=\frac{\omega }{2\pi }=\frac{2271}{2\times 3.14}=361.64Hz[/tex]