Question

A whistle emitting a sound of frequency 440Hz is tied to a string of length 1.5m and rotated with an angular velocity of 20 rad/s in the horizontal plane. Then, the range of frequencies heard by an observer stationed at a large distance from the whistle will be:
($${V_s} = 330m{s^{ - 1}}$$)
E.400.0 Hz to 484.0 Hz
F.403.3 Hz to 480.0 Hz
G.400.0 Hz to 480.0 Hz
H.403 Hz to 484.0 Hz

Answer 1

Hint- Use doppler effect to solve the sum which refers to change in wave frequency.


Complete step by step solution:
Given:
Frequency of the whistle; n=440Hz
Length of the string; L=1.5m
Angular velocity of the string; w=20rad/s
Consider the whistle as the source of linear velocity,
Thus, linear velocity of the source (whistle) is,
${v_{source}} = v = Lw = 1.5 \times 20 = 30m/s$
Let the frequencies heard by the observer from the source S’ and S be n’ and n’’.
Now we need to use Doppler effect in physics to solve the sum
So, before that let’s try and understand what doppler effect is
Doppler effect basically refers to the change in wave frequency during the relative motion between two things, firstly a wave source and secondary the observer. We can take an example to understand this better. When a sound object moves towards a person, the frequency of sound waves increases thus leading to a higher pitch.
Thus, now we solve the sum.
Doppler effect when source S’ approaches the stationary observer:
$n' = n\left[ {\dfrac{{{v_{sound}}}}{{{v_{sound}} - {v_{souce}}}}} \right] = 440\left[ {\dfrac{{330}}{{330 - 30}}} \right] = 484\,Hz$
Now we must find the doppler effect when the source approaches a stationary observer.


Doppler effect when source S’ approaches the observer is:
$n'' = n\left[ {\dfrac{{{v_{sound}}}}{{{v_{sound}} + {v_{souce}}}}} \right] = 440\left[ {\dfrac{{330}}{{330 + 30}}} \right] = 403\,Hz$
Thus, now we can find the frequency range.
Therefore,
The frequency range heard by the observer at a large distance is;

Option (d) i.e., 403 Hz to 484 Hz

Note: Students often do not understand the meaning of the question and they are not able to solve the sum. Students need to understand the meaning of the Doppler effect in physics to be able to solve the sum without any complications.