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# A sounding source at rest produces a note of frequency 350 Hz. If for an observer moving towards the source, the frequency appears to be 375 Hz, what is his speed? What will be the frequency for the observer when moving away from the source at the same speed? Velocity of sound = 320 m/s.

Last updated date: 23rd Jun 2024
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Hint: We are given a stationary source. In the first case, the receiver is moving towards the stationary source. We can find the velocity of the observer from the doppler’s effect on this situation. Once we have the velocity of the observer, consider that observer is moving away from the source with the same velocity. From the respective formula of doppler’s effect in this situation, we can have the frequency observed.
Formula used:
\eqalign{ & f = {f_0}\left( {\dfrac{{v + {v_r}}}{v}} \right) \cr & f' = {f_0}\left( {\dfrac{{v - {v_r}}}{v}} \right) \cr}

In the question, it is given that the source is initially at rest. And the receiver is in motion.
Let us consider the first case where the receiver is moving towards the source. According to the doppler effect when the source is stationary and the receiver is moving towards the source, the apparent frequency is given by
$f = {f_0}\left( {\dfrac{{v + {v_r}}}{v}} \right)$
Where,
f is the apparent frequency;
$f_0$ is the actual frequency of sound;
v is the velocity of sound in the medium; and
$v_r$ is the velocity of the receiver.
We have the actual frequency to be 350Hz, the apparent frequency as 375Hz, and the velocity of the sound is given as 320 m/s. Substituting these values in the formula, we get
\eqalign{ & f = {f_0}\left( {\dfrac{{v + {v_r}}}{v}} \right) \cr & \Rightarrow 375 = 350\left( {\dfrac{{320 + {v_r}}}{{320}}} \right) \cr & \Rightarrow 320 + {v_r} = \dfrac{{375 \times 320}}{{350}} \cr & \Rightarrow {v_r} = \dfrac{{375 \times 320}}{{350}} - 320 = 22.85m/s \cr & \therefore {v_r} = 22.85m/s \cr}
Therefore, the velocity of the observer is 22.85 m/s.
Now, let us consider the case where the observer is moving away from the source which is stationary. If the apparent frequency, in this case, in this case, is, say, $$f'$$. From doppler’s effect, the formula for $$f'$$ is given by
$f' = {f_0}\left( {\dfrac{{v - {v_r}}}{v}} \right)$
Substituting the values, we get
\eqalign{ & f' = {f_0}\left( {\dfrac{{v - {v_r}}}{v}} \right) \cr & \Rightarrow f' = 350\left( {\dfrac{{320 - 22.5}}{{320}}} \right) \cr & \Rightarrow f' = 350 \times 0.93 = 325.5Hz \cr & \therefore f' = 325.5Hz \cr}
Therefore, the frequency heard by the observer while moving away from the stationary source is 325.5 Hz.

Note:
Doppler effect is the change in frequency observed during the relative motion of the source and its observer, i.e., the frequency may increase or decrease with respect to the relative motion. The overall formula for the doppler effect can be given by $f = {f_0}\left( {\dfrac{{v \pm {v_r}}}{{v \pm {v_s}}}} \right)$. Where their true meanings hold as in the problem. The doppler’s effect formula for each case may be hard to remember. So, you can keep this formula in mind and change it with respect to the condition. If the receiver is moving towards the source, the numerator variables are added, and vice versa. If the source is moving towards the receiver the variable will add up in the denominator, vice versa.