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) \[200Hz\]
B) \[202Hz\]
C) \[196Hz\]
D) \[204Hz\]

Answer 1

Hint: Recall the concept of waxing and waning of tuning forks. When something is loaded on a tuning fork its frequency increases. As the mass of the fork increases its velocity decreases.

Complete step by step solution:
We now know that when mass is added to a fork, its velocity decreases and beat frequency increases. Let \[{{n}_{0}}\] be the frequency of fork \[1\] and \[n\] be the frequency of fork \[2\] in the initial state.
Now the difference of the frequencies initially is given to be \[6\]. Thus we have the equation:
\[{{n}_{0}}-n=4\]
\[\Rightarrow n={{n}_{0}}-4\]
\[\Rightarrow n=200-4\]
\[\Rightarrow n=196\]

Hence the original frequency of fork \[2\] was \[196Hz\].

Additional Information: Tuning forks work by releasing a nearly perfect wave pattern. It is claimed that God’s frequency is \[39.17MHz\] . The tuning fork works on the principle based on the changes of vibration frequency of the tuning fork when it comes into contact with a liquid or solid material. Tuning forks contain piezoelectric crystals built into the vibration tube that produces vibrations/resonations at certain frequencies. A tuning fork shows us how a vibrating object can produce sound. The fork consists of a handle and two prongs. When the tuning fork is hit with a rubber hammer, the prongs begin to vibrate. The back and forth vibration of the prongs produce disturbances of surrounding air molecules when as a result produces sound.

Note: Property of stationary waves, All particles except nodes perform S.H.M. During the formation of stationary waves the medium is broken into equally spaced loops.