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:

Hi Dear Sangram Patra

The frequency of nth harmonic is given by

v_n=(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!

k = 4L/(2n+1)                     (k is wavelength)

for n=0 ,  k =  4L = 8              (fundamental)

for n=1 , k = 4L/3   = 8/3          (first overtone)

for n = 2 , k =  4L/5 = 8/5             (second overtone)

for n =3 ,  k = 4L/7  = 8/7          (thiird overtone)

f (frequency) = v/k                  v = speed of light  

ratio of lower overtone (2nd overtone) & fundamental = f₂/f₀ = K₀/K₂ 8/(8/5) = 5:1