Question
Question: Two tuning forks P and Q sounded together and 6 beats per second are heard. P is in unison with a 30...
Two tuning forks P and Q sounded together and 6 beats per second are heard. P is in unison with a 30 cm air column open at both ends and Q is in resonance when the length of the air column is increased by 2 cm. The frequencies of forks P and Q are
A.) 90 Hz and 84 Hz
B.) 100 Hz and 106 Hz
C.) 96 Hz and 90 Hz
D.) 206 Hz and 200 Hz
Solution
Hint: We will start solving this question by finding the frequencies of the two tuning forks given, then their difference will give us the number of beats per second. Upon equating the difference in frequencies with the given number of beats, we will get the velocity of both tuning forks and with which we can find their actual frequencies.
Formula used:
Frequency of a tuning fork, f=2lv, where v is the velocity of sound wave generated by the tuning fork and l is the length of air column tuning fork .
Complete step by step solution:
We have been given two tuning forks P and Q who when sounded together, 6 beats per second is heard.
Length of the air column of the tuning fork P is 0.30 m and it starts resonating with Q when tuning fork Q has 0.32 m of air column.
We know that frequency of a tuning fork, f=2lv, where v is the velocity of sound wave generated by the tuning fork and l is the length of air column tuning fork is in unison with.
Therefore, frequency of tuning fork P and Q can be given by fP=2×0.30vHz and fQ=2×0.32vHz
According to question, upon resonance 6 beats per second is heard means fP−fQ=6
⟹2×0.30v−2×0.32v=6
⟹v=0.32−0.302×0.30×0.32×6=57.6ms−1
Thus, frequency of tuning fork P, fP=2×0.3057.6=96Hz
And, frequency of tuning fork Q, fQ=2×0.3257.6=90Hz
Hence, option c is the correct answer.
Note: Two different waveforms are generated when the tuning forks start to sound together. Betas are the interference points between the two tuning forks. When the frequencies of the two tuning forks are the same, the waves show a doubled amplitude.