Question

A tuning fork arrangement pair produces 4 beats/second with one fork of frequency 288 cps. A little wax is placed on the unknown fork and then produces 2 beats/s. The frequency of the unknown fork is?
A: 286 Hz
B: 292 Hz
C: 294 Hz
D: 288 Hz

Answer 1

Hint: Tuning fork is an instrument widely used in laboratories to explain the vibrations and the production of the sound. The intensity of the force applied determines the frequency of the sound produced by the string. In one second the number of times tuning fork oscillates is known as its frequency.

Complete step by step answer:
Initially we have 4 beats per second, we know beats are determined by the difference in the two given frequencies. So, one thing is clear that initially either the frequency is 292Hz or 288 Hz.
When we apply wax on the tuning fork, the frequency gets decreased. SO, the number of beats per second gets reduced.
$\Rightarrow f-288=4 \\
\Rightarrow f=288+4 \\
 \therefore f=292Hz \\ $

So, the correct answer is “Option B”.

Additional Information:
when two sound waves having different frequencies, comes nearer to us, the alternate constructive and destructive interference takes place. Interference means redistribution of energy either becoming maxima or minima. Due to this interference, sound sometimes becomes soft and other times loud. This phenomenon is called beats.
Note:
The standard unit of measuring the frequency is per second or Hertz. The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In order to find beat frequency, we use the following formula: \[{{f}_{beat}}=\left| {{f}_{2}}-{{f}_{1}} \right|\]
The number of beats per second is equal to the difference in frequency.