Question

When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200Hz, then what was the original frequency of fork 2?
A. 202Hz
B. 200Hz
C. 204Hz
D. 196Hz

Answer 1

Firstly, we should understand that the frequency will be decreased when the prong of a tuning fork is taped. But as the number of beats heard on simultaneous sounding of the given two forks is increased after the taping, we could conclude that the frequency of the first fork is greater.

Complete answer:
In the question, we are given two tuning forks with frequencies ${{f}_{1}}$ and ${{f}_{2}}$ respectively. We are also given the frequency of the first fork as 200Hz. When both forks are sounded simultaneously number of beats heard per second are,
$n=4$ …………………………………… (1)
Now when the tuning forks are sounded simultaneously with the prong of the second fork attached with some tape, number of beats heard are,
$n'=6$
${{f}_{1}}'-{{f}_{2}}'=6$
But we know that by attaching some tape on the prong of the second fork, its frequency is actually reduced. So, as the number of beats heard is increasing after attaching the tape on the second fork, we could understand that the frequency of the first fork is greater than the second. So, from (1) we have,
${{f}_{1}}-{{f}_{2}}=4$
But,
${{f}_{1}}=200Hz$
$\Rightarrow {{f}_{2}}=200-4$
$\therefore {{f}_{2}}=196Hz$
Therefore, we found the original frequency of fork 2 to be 196Hz.

Hence, option D is the right answer.

Note:
What we should know here is that beat frequency is given by the difference in the frequencies of the notes produced by the two tuning forks that interfere to produce the beats. Beats by definition is also the periodically repeating fluctuations in the intensity of the sound heard as the result of this interference. No sound is heard when destructive interference occurs.