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sangram patra 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:

6 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!


 


 


 

6 years ago
vikas askiitian expert
510 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

6 years ago
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