Test Index

Waves

Section: Physics

© examsnet.com

Question : 13 of 31

Marks: +1, -0

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0m/s and the man behind walks at a speed 2.0m/s . A third man is standing at a height 12m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330m/s . At the instant, when the moving men are 10m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is

[JEE Adv 2018 P1]

Your Answer:

Solution:

f_A = 1430\left[ \frac{330}{330 - 2\cos\theta} \right] = 1430\left[ \frac{1}{1 - \frac{2\cos\theta}{330}} \right] = 1430\left[ 1 + \frac{2\cos\theta}{330} \right]

f_B = 1430\left[ \frac{330}{330 + \cos\theta} \right] = 1430\left[ 1 - \frac{\cos\theta}{330} \right]

\Delta f = f_A - f_B = 1430\left[ \frac{3\cos\theta}{330} \right] = 13\cos\theta = 13\left( \frac{5}{13} \right) = 5.00

Hz

© examsnet.com

Go to Question:

Prev Question

Next Question