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A source and an observer approach each other with same velocity 50 m/s. If the frequency

A Source And An Observer Approach Each Other With Same Physics Question

A source and an observer approach each other with same velocity 50 m/s. If the frequency is 435 s⁻¹, then the real frequency is (Velocity of sound=332 m/s)

Options

(a) 320 s⁻¹
(b) 360 s⁻¹
(c) 390 s⁻¹
(d) 420 s⁻¹

Correct Answer:

320 s⁻¹

Explanation:

Both the source and the observer are approaching each other with the same velocity. v = 332 m/s v(o) = v(s) = 50 m/s Therefore, n’ = n.[v + v(o)] / [v – v(s)] 435 = n.[(332 + 50) / (332 – 50)] = (n × 382) / 282 n = (435 × 282) / 382 ≈ 321 ≈ 320 s⁻¹

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Topics: Waves (80)
Subject: Physics (2479)

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2 Comments on A source and an observer approach each other with same velocity 50 m/s. If the frequency

  1. When source is moving towards stationary object
    F’=(v/v-vs)f
    When source is stationary and observer is moving
    F’=(v+v0/v)f
    When both are moving towards each other
    F’=(v+v0/v-vs)f
    435=(330+50/330-50)f
    435=(380/280)f
    f=435×280/380
    f=435×14/19
    f=320.52
    f=320s

  2. When source is moving towards stationary object
    F’=(v/v-vs)f
    When source is stationary and observer is moving
    F’=(v+v0/v)f
    When both are moving towards each other
    f=(v+vo/v-vs)f
    435=(330+50/330-50)f
    435=(380/280)f
    f=435×280/380
    f=435×14/19
    f=320.52
    f=320s

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