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The maximum particle velocity in a wave motion is half the wave velocity. Then the amplitude of the wave is equal to
Options
(a) λ/4π
(b) 2λ/π
(c) λ/2π
(d) λ
Correct Answer:
λ/4π
Explanation:
For a wave, y = a sin [(2πvt/λ) – (2πx/λ)]
Here v = velocity of wave
.·. y = a sin [(2πvt/λ) – (2πx/λ)]
dy/dt = a (2πv/λ) cos [(2πvt/λ) – (2πx/λ)]
velocity = (2πav/λ) cos [(2πvt/λ) – (2πx/λ)]
Maximum velocity is obtained when
cos [(2πvt/λ) – (2πx/λ)] = 1
.·. v = (2πav/λ)
Then, v = v/2
(2πav/λ) = v/2 or a = λ/4π.
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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The wavelength of the short radio waves, microwaves, ultraviolet waves are λ₁,λ₂ and λ₃
- The circular motion of a particle with constant speed is
- In insulators (CB is Conduction Band and VB is Valence Band)
- Magnetic flux φ (in weber) linked with a closed circuit of resistance 10 Ω varies
- X-rays when incident on a metal
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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Since max. Particle velocity= aω
Wave velocity = ω/k
Acc to que
aω = 1/2 (ω/k)
Solving ,
a = 1/2k
Since k= 2π/ λ
Hence a = λ/4π