
Two sources of sound S1 and S2 produce sound waves of the same frequency 660 Hz. A listener is moving from source S1 towards S2 with constant speed u $\dfrac{m}{s}$ and he hears10 beats per second. The velocity of sound is $330 \dfrac{m}{sec}$ then u is equal to:
$\begin{align}
& A.10.0\dfrac{m}{s} \\
& B.5.5\dfrac{m}{s} \\
& C.15\dfrac{m}{s} \\
& D.2.5\dfrac{m}{s} \\
\end{align}$
Answer
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Hint: Use the Doppler effect concept and find the value of the apparent frequencies of the two sources found by the observer and find the difference of the apparent frequencies which is equal to the number of beats heard by the observer
Complete step-by-step answer:
The observer hears a different frequency from what the source produces as explained by the doppler effect. Here we need to find the apparent frequencies of both the sources found by the person as he moves with the speed u
The apparent frequency of source 1 heard when the person is moving towards Source 2 is given by
$n_1 = (\dfrac{v-v_o}{v-v_s})n$ where the n is the true frequency and the $v_o$ is the velocity of the observer and the $v_s$ is the velocity of the source moving.
So we get the apparent frequency of the source 1 as heard by the man is
$n_1 = (\dfrac{v-v_o}{v-v_s})n = (\dfrac{330-u}{330+0})660 = (330-u)(2)$
The apparent frequency of source 2 heard when the person is moving towards Source 2 is given by
$n_1 = (\dfrac{v-v_o}{v-v_s})n$
So we get the apparent frequency of the source 2 as heard by the man is
$n_2 = (\dfrac{v-v_o}{v-v_s})n = (\dfrac{330+u}{330+0})660 = (330+u)(2)$
Now the number of beats is the difference in the apparent frequency heard by the man
So we get that
$n_2 - n_1 =10 = 2u +2u = 4u$
So we get $u = \dfrac{10}{4} = 2.5 \dfrac{m}{s}$
Note: The most common mistake that one tends to make in the problems related to the doppler effect the sign convention to be taken for the two sources depending on the direction of the movement of the observer. The sign is negative when the observer is moving away and the sign is positive when the observer is moving towards the source.
Complete step-by-step answer:
The observer hears a different frequency from what the source produces as explained by the doppler effect. Here we need to find the apparent frequencies of both the sources found by the person as he moves with the speed u
The apparent frequency of source 1 heard when the person is moving towards Source 2 is given by
$n_1 = (\dfrac{v-v_o}{v-v_s})n$ where the n is the true frequency and the $v_o$ is the velocity of the observer and the $v_s$ is the velocity of the source moving.
So we get the apparent frequency of the source 1 as heard by the man is
$n_1 = (\dfrac{v-v_o}{v-v_s})n = (\dfrac{330-u}{330+0})660 = (330-u)(2)$
The apparent frequency of source 2 heard when the person is moving towards Source 2 is given by
$n_1 = (\dfrac{v-v_o}{v-v_s})n$
So we get the apparent frequency of the source 2 as heard by the man is
$n_2 = (\dfrac{v-v_o}{v-v_s})n = (\dfrac{330+u}{330+0})660 = (330+u)(2)$
Now the number of beats is the difference in the apparent frequency heard by the man
So we get that
$n_2 - n_1 =10 = 2u +2u = 4u$
So we get $u = \dfrac{10}{4} = 2.5 \dfrac{m}{s}$
Note: The most common mistake that one tends to make in the problems related to the doppler effect the sign convention to be taken for the two sources depending on the direction of the movement of the observer. The sign is negative when the observer is moving away and the sign is positive when the observer is moving towards the source.
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