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. A spring gun of spring constant 90 N/cm is compressed 12 cm by a ball of mass
  2. A bullet comes out of the barrel of gun of length 2m with a speed of 800 m/s
  3. The force of repulsion between two electrons at a certain distance is F.
  4. Two circuits have a mutual inductance of 0.1 H. What average e.m.f. is induced
  5. Which radiations are used in tratement of muscles ache?