A simple harmonic wave train of amplitude $2cm$ and time period $0.01\sec $ is travelling with a velocity of $10m{s^{ - 1}}$ in the positive x direction. The displacement of the particle from the mean position, the particle velocity and particle acceleration at $x = 50cm$ from the origin and at $t = 3\sec $ are:
A) $0,0,0$
B) $0,400\pi ,0$
C) $0,0,400\pi $
D) $400\pi ,0,0$

Hint: We have to use equation of simple harmonic motion $y = A\sin \left( {\omega t - kx} \right)$.Using this equation find velocity and acceleration by differentiating it. Insert the given values in the equations then we have our solutions.

Complete step by step solution:
Here given values are amplitude (A) is $2cm$
Time period (T) is $0.01\sec $
Velocity is $10m{s^{ - 1}}$
Therefore $\omega = \dfrac{{2\pi }}{T}$ is equal to $\dfrac{{2\pi }}{{0.01}} = 200\pi $
Now for $k = \dfrac{{2\pi }}{\lambda }$ rotation about a point
After all this we have $vk = \omega $
$10\dfrac{{2\pi }}{\lambda } = \dfrac{{2\pi }}{{0.01}}$
$\lambda = 0.1m$
So from the simple harmonic motion equation $y = A\sin \left( {\omega t - kx} \right)$
Differentiating both sides
$v = \dfrac{{dy}}{{dt}} = A\omega \cos \left( {\omega t - kx} \right)$
By again differentiating
$a = \dfrac{{{d^2}y}}{{d{t^2}}} = A{\omega ^2}\sin \left( {\omega t - kx} \right)$
Now put all above values in the equation of acceleration
$ \Rightarrow 2{\left( {200\pi } \right)^2}\sin \left( {200\pi \times 3\sec - 20\pi } \right)$
$ \Rightarrow 2 \times 400\pi \times \sin \left( {600\pi - 20\pi } \right)$
Make angle of sine be the even multiple of $2\pi $
$ \Rightarrow 800\pi \times \sin \left( {29 \times 2\pi } \right)$
As $\sin 2\pi $is 0. Therefore
$ \Rightarrow 800\pi \times 0 = 0$
Hence acceleration is zero.
Now put values in velocity equation
$v = \dfrac{{dy}}{{dt}} = A\omega \cos \left( {\omega t - kx} \right)$
$ \Rightarrow 2 \times 200\pi \times \cos \left( {200\pi \times 3 - 20\pi } \right)$
$ \Rightarrow 2 \times 200\pi \times \cos \left( {600\pi - 20\pi } \right)$
Make angle of cos be the even multiple of $2\pi $
$ \Rightarrow 400\pi \times \cos \left( {29 \times 2\pi } \right)$
As $\cos 2\pi $is 1. Therefore,
$ \Rightarrow 400\pi \times 1 = 400\pi $
Hence velocity is $400\pi $.
Now for displacement
$y = A\sin \left( {\omega t - kx} \right)$
Put values in it
$ \Rightarrow 2 \times \sin \left( {200\pi \times 3 - 20\pi } \right)$
$ \Rightarrow 2 \times \sin \left( {600\pi - 20\pi } \right)$
Make angle of sine be the even multiple of $2\pi $
$ \Rightarrow 2 \times \sin \left( {29 \times 2\pi } \right)$
As $\sin 2\pi $ is 0. Therefore
$ \Rightarrow 2 \times 0 = 0$
Hence displacement is also zero.
Therefore displacement is 0, velocity is $400\pi $ and acceleration is also 0.

So, option (B) is correct.

Note: Always remember the differentiation of sin and cos. Don’t confuse them if it happens that all values should be wrong. Because at the same values there are different values present. Also when picking options see the correct order of quantities that should be given.