The maximum particle velocity in a wave motion is half the wave velocity.

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.

admin:

View Comments (1)

  • 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

Related Questions

  1. According to Hook’s law, force is proportional to
  2. The sum of the magnitudes of two forces acting at a point is 16 N
  3. When you make ice cubes, the entropy of water
  4. Before using the tangent galvanometer, its coil is set in
  5. Two balls are dropped from heights h and 2h respectively. The ratio of times