Question

# Two men are walking along a horizontal straight line in the same direction. The man in the front walks at a speed 2 m/s and the man behind walks at a speed 4 m/s. A third man is standing at a height 24 m above the same horizontal line such that all three men are in the same vertical plane. The two walking men are blowing identical whistles which emit sounds of frequency 2010 Hz. The speed of sound in air is 335 m/s. At the instant when the moving men are 14 m apart, the stationary man is equidistant from them. The number of beats per second heard by the stationary man is (give your answer as the nearest integer value)

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Solution

## When the stationary man is equidistant from the two moving men (diagram shown below): Given that, Actual frequency heard by the observer at rest when source is also at rest is given by f0=2010 Hz Distance, AB=14 m So, AP=PB=7 m Height (OP)=24 m In ΔOAP OA=√(AP)2+(OP)2=√72+242=25 m and, cosθ=725 Now for source A, frequency received by stationary observer fA=f0[VV−VAOcos θ] [where V is speed of sound and VAO is speed of A w.r.t observer O ] Substituting the given data, fA=2010⎡⎢ ⎢ ⎢⎣335335−4×725⎤⎥ ⎥ ⎥⎦≈2017 Hz similarly for source B, fB=f0[VV+VBOcosθ] [where V is speed of sound and VBO is speed of B w.r.t observer O ] Substituting the given data, fB=2010⎡⎢ ⎢ ⎢⎣335335+2×725⎤⎥ ⎥ ⎥⎦≈2007 Hz So, beat frequency heard by stationary observer f=fA−fB=2017−2007=10 Hz i.e 10 beats per second.

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