a whistle producing sound wave of frequency 9500 hz and above its approaching a stationary person with speed of v m/s velocity of sound in air is 300 m/s . if a person can here frequency upto 10000 hz . what is maximum value of v which hear whistle.

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Solution

$A p p a r e n t f r e q u e n c y, f_{p} = \frac{V e l o c i t y o f S o u n d i n a i r}{V e l o c i t y o f S o u n d i n a i r \pm v_{s o u r c e}} W h e r e p o s i t i v e a n d n e g a t i v e s i g n a r e f o r g o i n g a w a y a n d t o w a r d s r e s p e c t i v e l y . f_{p} = 10000 = \frac{300}{300 - v} \times 9500 v = 255 m / s e c$

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a whistle producing sound wave of frequency 9500 hz and above its approaching a stationary person with speed of v m/s velocity of sound in air is 300 m/s . if a person can here frequency upto 10000 hz . what is maximum value of v which hear whistle.

Apparent frequency ,fp=Velocity of Sound in airVelocity of Sound in air±vsourceWhere positive and negative sign are for going away and towards respectively.fp=10000=300300-v×9500v=255m/sec