Question

Under what condition, maximum particle velocity is four times the wave velocity corresponding to the equation, y=A sin(omegat-kx)

Answer 1

At A=4/K, maximum particle velocity is four times the wave velocity.

We know that maximum particle velocity:wA

Also wave velocity:w/K

So according to question wA=4w/K

Hence A=4/K

Answer 2

Given: A sinusoidal wave $y=A sin(\omega at-kx)$ is given.

To Find: Condition for the velocity, maximum particle velocity is four times the wave velocity.

Solution:

                 Maximum particle velocity $= \omega A$

                 Wave velocity $=\frac{\omega}{k}$

Now, We have given that

Maximum velocity of the particle = 4 times of wave velocity

Putting the value of maximum particle velocity and wave velocity

we get ,

                          $\omega A$   $= 4\frac{\omega}{k}$

                           $A= \frac{4}{k}$

from the above relation we can clearly see that when A is $\frac{1}{4}th$ times of k then the maximum particle velocity is four times the wave velocity.

So, this is the required condition.