99 Percentile Qs Bank for JEE MainPhysicsWaves and Sound
Two sources (A ) and (B ) are producing notes of frequency (680 ~Hz ). A listener moves from (A ) to (B ) with a constant velocity (v ). if the speed of sound in air is (340 ~ms ⁻¹ ), the value of (y ) so that he hears 10 beats per second is
Options
- A(2.0 ~ms ⁻¹ )
- B(2.5 ~ms ⁻¹ )
- C(3.0 ~ms ⁻¹ )
- D(3.5 ~ms ⁻¹ )
Correct answer
B. (2.5 ~ms ⁻¹ )
Step-by-step solution
Given, notes of frequency produced by the sources (A ) and (B ) is (680 ~Hz ). i. e., (f_A ) and (f_B=680 ~Hz ) Velocity of listener moves from (A ) to (B ) is constant (=v ), speed of sound, (v_s=340 ~m / s ), and beats per second, (n=10 ) Now, beats per second from point (A ) to (B ) is given as ( aligned n & =f_A ( v_s+u v_s )-f_B ( v_s-u v_s ) 10 & =680 ( 340+u 340 )-680 ( 340-u 340 ) 10 & =680 [ ( 340+u 340 )- ( 340-u 340 ) ] 1 & =68 [ 340+u-340+u 340 ] 1 & =68 ( 2 u 340 ) 2 u & = 340 68 = 340 68 2 u & =2.5 ~m