Question

Two tuning forks with natural frequencies \(340~\text{Hz}\) each move relative to stationary observer. One fork moves away from the observer while the other moves towards him at the same speed. The observer hears beats of frequency \(3~\text{Hz}\). Find the speed of the tuning fork (velocity of sound in air is \(340~\text{m/s}\)).

Answer 1

Given,
Velocity of sound, \(v=340~\text{m/s}\)
Natural frequency of tuning fork, \(f_0=340~\text{Hz}\)

Let,
\(f _1=\) frequency observed by observer due to fork \(1\),
\(f _2=\) frequency observed by observer due to fork \(2\),
\(v_s=\) velocity of tuning fork,


As we know that,
\(f_1=f_0\left ( \dfrac{v}{v-v_{s}} \right )\) and

\(f_2=f_0\left ( \dfrac{v}{v+v_{s}} \right )\)

According to the problem, the beat frequency,
\(f_{1}-f_{2}=3\)

Substituting the values we get,
\(f_0\left ( \dfrac{v}{v-v_{s}} \right )-f_0\left ( \dfrac{v}{v+v_{s}} \right )=3\)

\(\Rightarrow \left [ \dfrac{1}{\left ( 1-\dfrac{v_{s}}{v} \right )}-\dfrac{1}{1+\dfrac{v_{s}}{v}} \right ]f_0=3\)

\(\Rightarrow\left [ \left ( 1-\dfrac{v_{s}}{v} \right )^{-1}- \left ( 1+\dfrac{v_{s}}{v} \right )^{-1} \right ]f_0=3\)

By binomial expansion with neglecting higher terms,

\(\Rightarrow\left [ \left ( 1+\dfrac{v_{s}}{v} \right )- \left ( 1-\dfrac{v_{s}}{v} \right ) \right ]f_0=3\)

\(\Rightarrow\dfrac{2v_{s}f_0}{v}=3\)

\(\Rightarrow v_{s}=\dfrac{3v}{2f_0}\)

Substituting the values, we get

\(v_{s}=\dfrac{(3)(340)}{(2)(340)}=1.5~\text{m/s}\)

Hence, correct option is (a).