Question

In a transverse wave distance between crest and neighbouring trough at the same instant is 4 cm and the distance between crest and trough at the same place is 1cm the next crest appear at same place after an interval of 4s the maximum speed of vibrating particle

Answer 1

Given:

In a transverse wave distance between the crest and neighbouring trough at the same instant is 4 cm and the distance between crest and trough at the same place is 1cm the next crest appear at the same place after an interval of 4s  

To find:

The maximum speed of vibrating particle

Solution:

From given, we have,

In a transverse wave distance between a crest and neighbouring trough at the same instant is 4 cm.  

The next crest appear at the same place after an interval of 4s

The traverse wave equation is given by,

y = a sin (wt + kx)

The speed of the particle is given by,

v = dy/dt = aw sin (wt + kx)

v_{max} = aw

As the distance between crest and trough at the same place is 1cm, so we have,

a = 1/2 cm.

Thus we have,

v_{max} = 1/2 × 2π/T

v_{max} = 1/2 × 2π/4

v_{max} = π/4 cm/s

Therefore, the maximum speed of vibrating particle is π/4 cm/s