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. With the decrease of current in the primary coil from 2 amperes to zero value
  2. If two bodies are projected at 30⁰ and 60⁰ respectively with the same velocity, then
  3. A tuning fork gives 4 beats with 50 cm length of a sonometer
  4. The potential energy of a system increases if work is done
  5. A radioactive element forms its own isotope after 3 consecutive disintegrations.