First slide

Doppler

Question

A whistle emitting a sound of frequency 440 Hz is tied to a string of 1.5 m length and rotated with an angular velocity of 20 rad/sec in the horizontal plane. Then the range of frequencies heard by an observer stationed at a large distance from the whistle will (v = 330 m/s):

Moderate

A

400.0 Hz to 484.0 Hz
B

403.3 Hz to 480.0 Hz
C

400.0 to 480. 0 Hz
D

403.3 Hz to 484.0 Hz

Solution

Frequency heard by the observed will be maximum when the source is in the position D. In this case, source will be approaching towards the stationary observer, almost along the line of sight (as observer is stationed at a larger distance).

Similarly, frequency heard by the observer will be minimum when the source reaches at the position B. Now, the source will be moving away from the observer.
$n_{\min .} = \frac{v}{v + v_{s}} \times n = \frac{330}{330 + 1.5 \times 20} \times 440$
= $\frac{330 \times 440}{360} = 403.3 Hz$
$n_{\max .} = \frac{v}{v - v_{s}} \times n = \frac{330}{330 - 1.5 \times 20} \times 440$
$= \frac{330 \times 440}{300} = 484 Hz$

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