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A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beats per second. The beat frequency does not change if the length of the wire is changed to 2 l cm. the frequency of the tuning fork (in Hz) must be(a) 200(b) 210(c) 205(d) 215

sumit kumar , 11 Years ago
Grade Upto college level
anser 2 Answers
Arun

Last Activity: 4 Years ago

Fundamental frequency is:
n = 1/2l √T/m
n 1/l
assume for 20cm frequency is n1
and for 21 cm frequency is n2
So
n1/n2 = 21/20
then n1 = 21x and n2 = 20x
Let frequency nf for producing 5 beats so n1 > nf > n2
So n1 - nf = 5
21x - nf = 5 ....(i)
And nf - n2 = 5
nf - 20x = 5 .....(ii)
from (1) ans (2) we get x = 205 hz

Vikas TU

Last Activity: 4 Years ago

 
Let the frequency of tuning fork be N
As the frequency of vibration string∝1 / length of string For sonometer wire of length 20cm,frequency must be (N+5) and that for the sonometer wire of length 21 cm, the frequency must be (N−5) as in each case the tuning fork produces 5 beats/sec with sonometer wire.
Hence n1I1=n2I2
⇒(N+5)×20=(N−5)×21
⇒N=205Hz

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