
Two engines pass each other moving in opposite directions with uniform speed of $30m/s$. One of them is blowing a whistle of frequency $540\;Hz$. Calculate the frequency heard by the driver of the second engine before they pass each other. Speed of sound is $330m/s$
\[\begin{align}
& A.648Hz \\
& B.540Hz \\
& C.270Hz \\
& D.450Hz \\
\end{align}\]
Answer
563.4k+ views
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 answer:
Let us consider the given sound wave to travel at a speed $c=330m/s$. Given that the speed of the observer and the source is $30m/s$, and that they move away from each other. Let us then assume that the speed of the engine A is $v_{A}=30m/s$, then clearly, the speed of the source with respect to the engine B is $v_{B}=-30m/s$, 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 $f=540Hz.$.
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 $f=\left(\dfrac{c+ v_{B}}{c-v_{A}}\right) f_{0}$, as both the source and observer are moving. Now substituting the values, we get $f=\left(\dfrac{330-30}{330+30}\right)f_{0}$
Or, $f_{0}=540\left(\dfrac{360}{300}\right)$
Or $f_{0}=648Hz$
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 answer:
Let us consider the given sound wave to travel at a speed $c=330m/s$. Given that the speed of the observer and the source is $30m/s$, and that they move away from each other. Let us then assume that the speed of the engine A is $v_{A}=30m/s$, then clearly, the speed of the source with respect to the engine B is $v_{B}=-30m/s$, 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 $f=540Hz.$.
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 $f=\left(\dfrac{c+ v_{B}}{c-v_{A}}\right) f_{0}$, as both the source and observer are moving. Now substituting the values, we get $f=\left(\dfrac{330-30}{330+30}\right)f_{0}$
Or, $f_{0}=540\left(\dfrac{360}{300}\right)$
Or $f_{0}=648Hz$
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

