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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
Related Questions: - Two metal rods 1 and 2 of same lengths have same temperature difference
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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two metal rods 1 and 2 of same lengths have same temperature difference
- After 300 days, the activity of a radioactive sample is 5000 dps
- For high frequency a capacity offers
- If dimensions of critical velocity of a liquid flowing through a tube are expressed
- What is the energy released by fission of 1 g of U²³⁵?
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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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
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