Question

An elastic string of length 2 m is fixed at its end. The string starts to vibrate in third overtone with a frequency 1200 Hz. The ratio of frequency of lower overtone and fundamental is

Answer 1

Answer:

Hey!

The frequency of nth harmonic is given by

vn=(2n+1)(T/u)1/2/4l

Given that it is 3rd overtone means 4th harmonic

so putting n=4

v4=9(T/u)1/2/4l=1200.............(1)

For lower overtone n=2

v2=5(T/u)1/2/4l ....................(2)

For fundamental n=1

v1=3(T/u)1/2/4l..................(3)

Dividing (1) by (2)

1200/v2=9/5 ...............(4)

Diving (1) by (3)

1200/v1=9/3 ....................(5)

Dividing (5) by (4)

v2/v1=5/3

This is the answer.

Thank you!❤️