Stationary wave is setup in a uniform string clamped at both the ends. Length of the string is 0.3 m.Snapshot of the string is taken the two instants one at t=0 sec and another at t=0.2 sec. These is two snapshots are shown below.

Velocity of point P (which is also the mid point of the string ) is in upward direction (take upward direction to be positive ) at t=0 sec.At the instant snapshots are taken particles are at half of their respective maximum displacement from mean position.During this time interval particles have crossed their mean position only once.Answer the following 3 questions for the qiven situation.
Velocity of travelling wave in the string is :

Options

A. 1 m/s
B. 0.5 m/s
C. 2 m/s
D. 0.25 m/s

Explanation

Displacement equations of point Q=`A sin (omegat+(5pi)/6)`
Equation of standing wave y(x)=A(x) `sin (omegat+(5pi)/6)=A sin kx.sin (omegat+(5pi)/6)`
According to snapshots
`t=1/5=pi/omega rArr omega =5pi` rad/s
Time period `T=(2pi)/(5pi)=2/5sec`
wavelength `lambda=0.2 m`
wave velocity `v=lambda/T= 2/10. 5/2 =1/2` m/s
Disp. equation for point P `y=A sin (omegat+(11pi)/6)`
velocity equation for point P `V_p=omegaA cos (omegat+(11pi)/6)`
Acceleration equation for point P `a_p=-omega^2A sin (omegat+(11pi)/6)`
here `omega=5pi` rad/s A=2cm