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```        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

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
Thank you!

```
6 years ago
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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