Question

P, Q and R are three tuning forks. Frequency of tuning fork P is $380\text{Hz}$. Number of beats produced by tuning forks P and Q is $10\text{Hz}$ and by tuning forks Q and R is $8\text{Hz}$. When wax is put on P, beat frequency produced by tuning forks P and Q becomes $4\text{Hz}$ and by tuning forks P and R becomes $12\text{Hz}$. Frequency of fork Q is greater than the frequency of fork R. Find natural frequency of tuning forks Q and R respectively.

• A. $380\text{Hz},370\text{Hz}$
• B. $370\text{Hz},378\text{Hz}$
• C. $378\text{Hz},380\text{Hz}$
• D. $370\text{Hz},362\text{Hz}$

Answer 1

The correct option is B $370\text{Hz},378\text{Hz}$

Let $f_1,f_2$ and $f_3$ be the frequencies of forks P, Q and R respectively.

Given:

$f_1=380\text{Hz}$

$|f_1-f_2|=10\text{Hz}$

$|f_2-f_3|=8\text{Hz}$

When fork P is waxed:

Its frequency will decrease and given that beat frequency between P and Q is also decreasing.

$\therefore f_1>f_2$

$\Rightarrow f_1-f_2=10\text{Hz}$

$\Rightarrow f_2=f_1-10=380-10=370\text{Hz}$

So,

$|370-f_3|=8\text{Hz}$

Given, $f_2>f_3$

$\therefore f_3=362\text{Hz}$

Hence, option $\left(B\right)$ is the correct answer.