Question

Two tuning forks have frequency $450Hz$and $454Hz$respectively. On sounding these forks together, the time interval between successive maximum intensities will be
(A) $\dfrac{1}{2}\sec $
(B) $\dfrac{1}{4}\sec $
(C) $1\sec $
(D) $2\sec $

Answer 1

Hint
Since we are given ‘nearly’ equal frequencies of $450Hz$and $454Hz$, the question is indirectly asking about the beat period.
$\Rightarrow T = \dfrac{1}{{{n_1} - {n_2}}}$

Complete step by step answer
In the given question the two tuning forks have nearly the same frequency, hence it results in formation of beats. The time interval between successive intensities is the beat period which is given by the equation,
$\Rightarrow T = \dfrac{1}{{{n_1} - {n_2}}}$, here T is the beat period and ${n_1}$,${n_2}$ are the frequencies of the two forks
Substituting the values of frequencies from the question in the above equation we get,
$\Rightarrow T = \dfrac{1}{{454 - 450}} = \dfrac{1}{4}\sec $
Hence the time between successive intensities is $\dfrac{1}{4}\sec $
Correct option is (B). $\dfrac{1}{4}\sec $.

Additional Information
Interference of two waves means that in a medium two waves of same frequency are travelling simultaneously in the same direction, then the resultant intensity is different than the sum of the two waves intensities.Interference can be ‘constructive’ if the resultant wave intensity at the point is maximum, while it is said to be ‘ destructive’ at a point if the resultant intensity is zero or minimum.

Note
Beat is an example of interference of sound. When two sound waves having nearly equal frequency have been produced. Due to superposition, the resultant sound wave intensity will increase and decrease alternatively with time. This increase and decrease in intensity are known as the phenomenon of beats. In one second, the number of times there is rise and fall observed in intensity is ‘beat frequency’. If the two frequencies are equal and they are sounded together, then we won’t observe rise and fall in intensity of resultant waves.