A man stands in front of a cliff and finds that when he claps at a regular interval of 1.2 seconds, he does not hear the echo distinctly. Then he moves closer to the cliff by 32 m and finds that he cannot hear the echo distinctly when he claps at an interval of 1 second. Find the velocity of the sound waves.
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$T h e p e r s o n d o e s n o t h e a r t h e s o u n d d i s t i n c t l y a s w h e n t h e e c h o r e a c h e s h i m a t t h e s a m e i n s \tan t h e c l a p s f o r t h e n e x t t i m e . L e t t h e p e r s o n s \tan d s a t a d i s \tan c e o f x f r o m t h e c l i f f . t i m e t a k e n f o r t h e s o u n d t o t r a v e l t o t h e c l i f f a n d r e t u r n t o t h e c l i f f a g a i n i s t_{1} = \frac{x}{v} + \frac{x}{v} = \frac{2 x}{v} (v i s t h e v e l o c i t y o f s o u n d) 1.2 = \frac{2 x}{v} . . . 1 x = \frac{1.2 v}{2} . . . a w h e n t h e p e r s o n m o v e s a d i s \tan c e 32 m t h a n h i s d i s \tan c e f r o m t h e c l i f f i s (x - 32) m t i m e t a k e n f o r t h e s o u n d t o t r a v e l t o t h e c l i f f a n d r e t u r n t o t h e c l i f f a g a i n i s t_{2} = \frac{x - 32}{v} + \frac{x - 32}{v} = \frac{2 (x - 32)}{v} 1 = \frac{2 (x - 32)}{v} . . . 2 x - 32 = \frac{v}{2} x = 32 + \frac{v}{2} . . . b e q u a t i n g a a n d b \frac{1.2 v}{2} = 32 + \frac{v}{2} (0.6 - 0.5) v = 32 v = \frac{32}{0.1} = 320 m / s R e g a r d s$

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