First slide

Doppler

Question

A whistle emitting a sound of frequency 440 Hz is said to a string of 1 . 5 m length and rotated with an angular velocity of20 rad,/s in horizontal plane. Then the range of frequencies heard by an observer stationed at a large distance from the whistle will be (v = 330 m/s)

Difficult

A

400 . 0 Hz to 484.0 Hz
B

4o3 . 3 Hz to 480 .0 Hz
C

4O0 .0 Hz to 480 .o Hz
D

403 . 3 Hz to 484.0 Hz

Solution

see fig
The observer will hear the frequency (maximum) when the source is in position D because source is moving towards observer almost along the line of sight. So,
$n_{max} = (\frac{V}{V - V_{s}}) n = (\frac{330}{330 - 1 \cdot 5 \times 20}) 440 = \frac{330}{300} \times 440 = 484 Hz$
The observer will hear the frequency (minimum) when the source is in position I because the source is moving away from the observer. So,
$n_{min} = (\frac{v}{v + v_{s}}) \times n = (\frac{330}{330 + 1 \cdot 5 \times 20}) \times 440 = \frac{330}{360} \times 440 = 403 \cdot 3 Hz$

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