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Question

A tuning fork of frequency 328 hz is moved towards a wal at a speed of 2 m/s .an observer standing on the same side as the fork hears two sounds one directly from the fork and the other reflectd from the wal.number of beats per second is (velocity of sound in air is 33
m/s)

Ashish Kumar · Accepted Answer

The apparent frequency observed of the sound directly coming from the tuning fork ,

f1=v/(v-Us) f, As the source is moving away from the observer Us will be negative.

f1= 330/(330 + 2) 328 = 326.02 Hz

The apparent frequency received at the wall will be F=v/(v -Us), for this the Us will be positive as it is moving towards the wall. therefore, F = 330 / (330 - 2) 328 = 330 Hz.

Now the wall will reflect this wave at frequency F and therefore act as a soure for the observer.As there is no relative motion between the source(wall) and the observer the freq osbserved for the reflected sound wave will be F=330 Hz

frequency of beats = F - f1 = 330 - 326.02 = 3.98 Hz

Wave Motion> anser-this-plz...

A tuning fork of frequency 328 hz is moved towards a wal at a speed of 2 m/s .an observer standing on the same side as the fork hears two sounds one directly from the fork and the other reflectd from the wal.number of beats per second is (velocity of sound in air is 33
m/s)

SRI CHAKRA KUMAR , 12 Years ago

Ashish Kumar

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Wave Motion> anser-this-plz...

A tuning fork of frequency 328 hz is moved towards a wal at a speed of 2 m/s .an observer standing on the same side as the fork hears two sounds one directly from the fork and the other reflectd from the wal.number of beats per second is (velocity of sound in air is 33m/s)

SRI CHAKRA KUMAR , 12 Years ago

Ashish Kumar

LIVE ONLINE CLASSES

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