A source and an observer move away from each other with speed of $10m/s$ with respect to the ground. Apparent frequency of the source is \[1950Hz\]. The natural frequency of the source is (assume velocity of sound to be $340m/s$)
\[\begin{align}
& A.2068Hz \\
& B.1832Hz \\
& C.1950Hz \\
& D.1650Hz \\
\end{align}\]
Answer
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Hint: Doppler Effect is the change in frequency when the position of the observer changes with respect to the source. However, here we are assuming that the velocity of the wave is constant during the interaction. Also, the wave is either approaching the observer or moving away from the observer, only.
Formula used:
$f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$, where, $f$ is the apparent frequency of the sound, $f_{0}$ is the actual or real frequency of the sound, $c$ is the speed of the sound wave and $v_{s},v_{o}$ is the speed of the moving source and observer respectively.
Complete step-by-step answer:
Let us consider the given sound wave to travel at a speed $c=340m/s$. Given that the speed of the observer and the source is $10m/s$, and that they move away from each other. Let us then assume that the speed of the observer is $v_{o}=10m/s$, then clearly, the speed of the source with respect to the observer is $v_{s}=-10m/s$, since the source and the object move in the opposite direction to each other.
Also given that the apparent frequency of the sound wave is $f=1950Hz.$.
From Doppler’s law we know $f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$
Here, the Doppler’s law will become $f=\left(\dfrac{c- v_{o}}{c-v_{s}}\right) f_{0}$, as both the source and observer are moving .
Now substituting the values, we get
$1950=\left(\dfrac{340-10}{340+10}\right)f_{0}$
$\Rightarrow f_{0}=1950\left(\dfrac{350}{330}\right)$
$\Rightarrow f_{0}=2068Hz$
So, the correct answer is “Option A”.
Note: Doppler law is used only when the speed of the source and the observer are both less than the speed of the sound wave in the medium. Then clearly the relative frequency of a moving source or observer or both, is lesser than the frequency of the sound wave. This law can be extended to any waves.
Formula used:
$f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$, where, $f$ is the apparent frequency of the sound, $f_{0}$ is the actual or real frequency of the sound, $c$ is the speed of the sound wave and $v_{s},v_{o}$ is the speed of the moving source and observer respectively.
Complete step-by-step answer:
Let us consider the given sound wave to travel at a speed $c=340m/s$. Given that the speed of the observer and the source is $10m/s$, and that they move away from each other. Let us then assume that the speed of the observer is $v_{o}=10m/s$, then clearly, the speed of the source with respect to the observer is $v_{s}=-10m/s$, since the source and the object move in the opposite direction to each other.
Also given that the apparent frequency of the sound wave is $f=1950Hz.$.
From Doppler’s law we know $f=\left(\dfrac{c\pm v_{o}}{c\pm v_{s}}\right) f_{0}$
Here, the Doppler’s law will become $f=\left(\dfrac{c- v_{o}}{c-v_{s}}\right) f_{0}$, as both the source and observer are moving .
Now substituting the values, we get
$1950=\left(\dfrac{340-10}{340+10}\right)f_{0}$
$\Rightarrow f_{0}=1950\left(\dfrac{350}{330}\right)$
$\Rightarrow f_{0}=2068Hz$
So, the correct answer is “Option A”.
Note: Doppler law is used only when the speed of the source and the observer are both less than the speed of the sound wave in the medium. Then clearly the relative frequency of a moving source or observer or both, is lesser than the frequency of the sound wave. This law can be extended to any waves.
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