Two tuning forks exhibit beats at a beat frequency of 3 Hz. The frequency of one fork is 256 Hz. Its frequency is then lowered slightly by adding a bit of wax to one of its prong. The two forks then exhibit a beat frequency of 1 Hz. Determine the frequency of second tuning fork.
(Ans: 253 Hz)



Data Given:

Let the two tuning forks are P and Q
Number of beats per second before loading = N = 3
Number of beats per second after loading = N' = 1
Frequency of tuning fork P = `\f_P` = 256 Hz

To Find:

Frequency of tuning fork P before loading  = `\f_Q` = ?
Frequency of tuning fork P after loading  = `\f'_Q` = ?

Solution:

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

`\f_P` - `\f_Q` = ± 3   

or 

`\f_Q` = `\f_P`  ± 3

It mean the frequency of the tuning fork Q is either

 `\f_Q` = `\f_P`  + 3  = 256 Hz + 3 = 259 Hz

or 

`\f_Q` = `\f_P`  - 3 = 256 Hz - 3 = 253 Hz


Similarly after loading 

`\f_P` - `\f'_Q`  = ± 1   

or 

`\f'_Q` = `\f_P`  ± 1

It mean the frequency of the tuning fork Q is either

`\f'_Q` = `\f_P`  + 1  = 256 Hz + 1 = 257 Hz

or 

`\f'_Q` = `\f_P`  - 1 = 256 Hz - 1 = 255 Hz

Let the 259 is the exact frequency of the tuning fork `\f_Q`. But when the tuning fork `\f_A` is loaded slightly its frequency start decreasing say 255 Hz, 254 Hz, .......... and beat frequency (difference in both tuning fork frequencies) will goes on increasing and will never be equal to 1 (according to  the given condition) so the answer 259 will go wrong. 

Now for taking the tuning fork `\f_Q`. frequency 253 Hz. When the tuning fork `\f_A` is loaded slightly its frequency start decreasing from 256 Hz to 255 Hz then 254 Hz, Etc. At 254 Hz difference in the two tuning frequencies will equal to 1 Hz. 

Hence 253 Hz is the exact frequency of the tuning fork `\f_Q` 


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