Question

A speeding motorcyclist sees traffic jam ahead of him. He slows to \[36\,kmh{{r}^{-1}}\]. He finds that traffic has eased and a car moving ahead of him at \[18\,kmh{{r}^{-1}}\]is honking at a frequency of 1392 Hz. If the speed of sound is \[343\,m{{s}^{-1}}\], the frequency of the honk as heard by him will be :
A. 1332Hz
B. 1372Hz
C. 1412Hz
D. 1454Hz

Answer 1

Hint: The concept of Doppler effect is required to solve this problem. The Doppler’s formula required to solve this problem is, \[f={{f}_{0}}[\dfrac{{{c}_{s}}+{{v}_{r}}}{{{c}_{s}}+{{v}_{s}}}]\], where ${{v}_{s}}$ is the velocity of the source and ${{v}_{r}}$ is the velocity of the receiver. ${{f}_{0}}$ is the frequency of the horn originally.

Complete step by step answer:
Let’s start by making a diagram of the problem.

From the question, we are given velocity of the motorcyclist is \[36\,kmh{{r}^{-1}}\], that is, \[{{v}_{m}}=36\,kmh{{r}^{-1}}\]. The velocity of the car is also moving along the same direction with a velocity of \[18\,kmh{{r}^{-1}}\], that is \[{{v}_{c}}=18\,kmh{{r}^{-1}}\]. The velocity of sound in air is given to be \[330\,m{{s}^{-1}}\], that is, \[{{c}_{s}}=343\,m{{s}^{-1}}\]. The horn honked by the car is of frequency 1392Hz, hence ${{f}_{0}}=1392Hz$.
To find the frequency of the horn as heard by the motorcyclist, we will require the Doppler effect. Let’s first understand what does the Doppler effect mean. This effect gives the updated frequency as seen by an observer when a source is moving or the receiver is moving or either both the receiver and the source is moving. Due to the movement of the receiver and(or) the source, a superposition of waves takes place that emerge from the source. Hence, updated frequency (f) is observed by the receiver. The formula of the updated frequency for source frequency $({{f}_{0}})$ and speed of speed of wave in air is $({{c}_{s}})$ for the waves being sound wave and it will be (c) for the waves being light wave. The velocity of the source is, ${{v}_{s}}$ and ${{v}_{r}}$ is the velocity of the receiver. Therefore the Doppler’s formula is: \[f={{f}_{0}}[\dfrac{{{c}_{s}}+{{v}_{r}}}{{{c}_{s}}+{{v}_{s}}}]\].
For the current case, we have ${{f}_{0}}$ is the frequency of the horn originally as 1392Hz. The velocity of sound in air, $({{c}_{s}}=343m{{s}^{-1}})$. The velocity of the receiver is that of the motorcyclist \[{{v}_{r}}={{v}_{m}}=36\,kmh{{r}^{-1}}\]. Converting the velocity from kmph to m/s, we will multiply the value of the velocity with 5/18. Therefore, \[{{v}_{r}}={{v}_{m}}=36\,kmh{{r}^{-1}}\Rightarrow {{v}_{m}}=36\,(\dfrac{5}{18})m{{s}^{-1}}\Rightarrow {{v}_{m}}=10m{{s}^{-1}}\].
Similarly for the velocity of the source, that is the car with velocity of \[{{v}_{s}}={{v}_{c}}=18\,kmh{{r}^{-1}}\]. Converting this velocity as well, we get, \[{{v}_{s}}={{v}_{c}}=18\,kmh{{r}^{-1}}\Rightarrow {{v}_{c}}=18\,(\dfrac{5}{18})m{{s}^{-1}}\Rightarrow {{v}_{c}}=5m{{s}^{-1}}\].
Hence, using the Doppler’s formula, we find \[f={{f}_{0}}[\dfrac{{{c}_{s}}+{{v}_{r}}}{{{c}_{s}}+{{v}_{s}}}]\Rightarrow f=(1392)[\dfrac{343+10}{343+5}]\Rightarrow f=(1392)[\dfrac{353}{348}]\Rightarrow f=1412Hz.\]

Hence, the correct answer is option C.

Note:
The Doppler’s formula is: \[f={{f}_{0}}[\dfrac{{{c}_{s}}+{{v}_{r}}}{{{c}_{s}}+{{v}_{s}}}]\]. Here the values of the velocity of the receiver and that of the source is both positive and negative depending upon the direction of the movement of the source and the receiver. Hence, if the receiver is moving towards the source, then the value of ${{v}_{r}}$ will be positive. If the receiver is static or not moving, then the value of ${{v}_{r}}$ would be zero and lastly, when the receiver is moving away from the source, then the value of ${{v}_{r}}$ will be negative.
Similarly, if the source is moving away from the receiver, then the value of ${{v}_{s}}$ will be positive. If the source is static or not moving, then the value of ${{v}_{s}}$ would be zero and lastly, when the source is moving towards the receiver, then the value of ${{v}_{s}}$ will be negative.