Question

A tuning fork of unknown frequency produces 4 beats per second when sounded with another tuning fork of frequency 254 Hz. The same number of beats per second are produced when the unknown tuning fork is loaded with wax. Find the unknown frequency before and after loading with wax.
A) 258 Hz
B) 254 Hz
C) 250 Hz
D) Can not be determined.

Answer 1

When a tuning fork is loaded with wax, the frequency of the tuning fork decreases. Here, the beats produced per second remain the same before and after loading wax into the unknown tuning fork. The beats produced per second refers to the difference between the frequencies of the two tuning forks.

Formula used:
The frequency of the beats produced is given by, ${f_{beat}} = {f_1} - {f_2}$ where ${f_1}$ and ${f_2}$ are the frequencies of the two sound waves.

Complete step by step answer.
Step 1: Mention the key points of the problem at hand.
The beat frequency when the two forks were sounded together is 4 beats per second.
When the tuning fork of unknown frequency is loaded with wax, the beat frequency remains the same as 4 beats per second.
The frequency of the second tuning fork is known to be ${f_2} = 254{\text{Hz}}$
Let ${f_1}$ be the unknown frequency of the first tuning fork.
Step 2: Using the expression for beat frequency, we can find the unknown frequency ${f_1}$ before and after loading.
The frequency of the beats produced is given by, ${f_{beat}} = {f_1} - {f_2}$ ------- (1)
where ${f_1}$ and ${f_2}$ are the frequencies of the two sound waves produced by the first and second tuning forks respectively.
From equation (1), we can express the unknown frequency as ${f_1} = {f_2} + {f_{beat}}$ .
But we know the frequency of a sound wave can be ${f_1} = {f_2} + {f_{beat}}$ -------- (2)
or ${f_1} = {f_2} - {f_{beat}}$ ------- (3).
Before loading

We consider the case where wax is not loaded onto the first tuning fork.
Substituting for ${f_2} = 254{\text{Hz}}$ and ${f_{beat}} = 4$ in equations (2) and (3) we get, ${f_1} = 254 + 4 = 258{\text{Hz}}$ or ${f_1} = 254 - 4 = 250{\text{Hz}}$
So the two possible frequencies before loading wax are ${f_1} = 258{\text{Hz}}$ or ${f_1} = 250{\text{Hz}}$ .
After loading
We now consider the case where wax is loaded onto the tuning fork. The number of beats produced remains the same. However, the frequency ${f_1}$ of the fork must decrease.
Now if ${f_1} = 250{\text{Hz}}$, then on loading, it would become ${f_1} < 250{\text{Hz}}$ .Then the beat frequency will be ${f_{beat}} = 254 - {f_1} = 4$. This suggests that ${f_1} = 250{\text{Hz}}$ which is impossible. So, the frequency of the tuning fork must not be 250 Hz i.e., ${f_1} \ne 250{\text{Hz}}$ .
Thus the frequency of the tuning fork is ${f_1} = 258{\text{Hz}}$ .

Hence the correct option is A.

Note: After loading the frequency of the tuning fork decreases i.e., ${f_1} < 258{\text{Hz}}$. And it can be obtained from the relation, ${f_{beat}} = 254 - {f_1} = 4$. This implies that ${f_1} = 250{\text{Hz}}$ i.e., the frequency of the fork has decreased to 250 Hz on loading. While calculating the beat frequency we subtract the smaller frequency from the bigger one so that the beat frequency is positive.