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 time graph of particle at mid point of the string (i.e. particle P)

Options

A.
B.
C.
D. None of these

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