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

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.