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 body projected vertically from the earth reaches a height equal to earth’s radius
  2. A 50Hz AC signal is applied in a circuit of inductance of (1/π)H and resistance 2100Ω
  3. A solid sphere, a hollow sphere and a ring are released from top of an inclined
  4. The work of 146 kJ is performed in order to compress one kilo mole fo gas adiabatically
  5. Which phenomenon best supports the theory that matter has a wave nature?