Question

Two tuning forks of frequency 520 Hz and 524 Hz are vibrated together. The time interval between successive maximum and minimum intensities is
A) $ 1/8\,s $
B) $ 1/4\,s $
C) $ 1/2\,s $
D) $ 1\,s $

Answer 1

Hint: When two tuning forks of such similar frequencies are vibrated together, they will generate beats. The time interval between the maximum and minimum intensities will depend on the frequencies of the beats generated.

Formula Used: We will be using the following formula here,
 $ f = {f_1} - {f_2} $ where $ {f_1} $ and $ {f_2} $ are the frequencies of the two tuning forks and $ f $ is the net frequency of the beats.

Complete step by step answer:
When two tuning forks are vibrated together, they will generate beats. The beats will have a frequency equal to the difference of the frequency of the two tuning forks.
Hence the net frequency of the beats will be
 $ f = {f_1} - {f_2} $ where $ {f_1} $ and $ {f_2} $ are the frequencies of the two tuning forks.
So we can calculate the frequency of the beats numerically as
 $ f = 524 - 520 = 4\,Hz $
This implies that 4 beats will be generated in one second. Now in a beat, there will be a maximum intensity of sounds followed by the minimum intensity of sound. Since 4 beats will be generated in one second, the interval between the maximum and minimum intensity will be the inverse of the beats frequency as
 $ T = \dfrac{1}{f} $
Substituting the value of $ f = 4\,Hz $ , we get
 $ T = \dfrac{1}{4} = 0.25\,s $
Hence the time duration between successive maximum and minimum amplitude will be $ 1/4\,s $ which corresponds to option (B).

Note:
Beats are generated when sound waves generated by tuning forks with two similar frequencies interfere with each other. This creates a pattern of maximum and minimum amplitude of the sound that we hear. The frequency of the beats generated tells us how many times the maximum and minimum amplitude will occur in one second. The time period between successive maximum intensities and minimum intensities will be the same.