Two tuning forks P and Q give 4 beats per second. On loading Q Lightly with wax, we get 3 beats per second. What is the frequency of Q before and after loading if the frequency of P is 512 Hz? (516 Hz, 515Hz) 



Given:

Number of beats per second before loading = N = 4
Number of beats per second after loading = N' = 3
Frequency of tuning fork P = fP = 512 Hz

To Find:

Frequency of tuning fork P before loading  = fQ = ?
Frequency of tuning fork P after loading  = f′Q = ?

Solution:

As the number of beats per second is equal to the difference in frequencies of the two sources (tuning forks), So, the frequency of the tuning fork before loading  are

fP - fQ = ± 4   

or 

fQ = fP  ± 4

It means the frequency of the tuning fork Q is either

 fQ = fP  + 4  = 512 Hz + 4 = 516 Hz

or 

fQ = fP  - 4 = 512 Hz - 4 = 508 Hz


Similarly after loading 

fP - f′Q  = ± 3   

or 

f′Q = fP  ± 3

It means the frequency of the tuning fork Q is either

f′Q = fP  + 3  = 512 Hz + 3 = 515 Hz

or 

f′Q = fP  - 3 = 512 Hz - 3 = 509 Hz

Since after loading the number of beats per second decreases so, frequencies of the tuning fork before and after are: 

before loading

 fQ  = 516 Hz ------------Ans. 1

and after loading 

 f′Q  = 515 Hz ---------------Ans. 2


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