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Grade: 12

                        

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:

9 years ago

Answers : (2)

Vinay Arya
37 Points
							

Hi Dear Sangram Patra

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!

 

 

 

9 years ago
vikas askiitian expert
509 Points
							

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

9 years ago
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