Question

A body which is being oscillates with SHM according to the equation $x=5.0\cos \left( 2\pi t+\pi \right)$. At time $t=1.5s$, its displacement, speed and acceleration respectively will be given as,
$\begin{align}
  & A.0,-10\pi ,+20{{\pi }^{2}} \\
 & B.5,0,-20{{\pi }^{2}} \\
 & C.2.5,+20\pi ,0 \\
 & D.-5.0,+5\pi ,-10{{\pi }^{2}} \\
\end{align}$

Answer 1

Hint: First of all find the displacement of the body by the substitution of the time taken in it. The velocity of the body can be found by taking the time derivative of the displacement of the same body. And substitute the time taken for the specific velocity of the body. The acceleration will be the time derivative of the velocity of the body. And we have to substitute the same time taken in order to get the acceleration of the body at that instant. This will help you in answering this question.

Complete step by step solution:
The displacement of the body has been given as,
$x=5\cos \left( 2\pi t+\pi \right)$
At the time mentioned as,
$t=1.5s$
Substituting this value in the equation will be given as,
\[\begin{align}
  & x=5\cos \left( 2\pi \times 1.5+\pi \right)=5\cos \pi \\
 & \Rightarrow x=5m \\
\end{align}\]
The velocity of the body is found by taking the derivative of the displacement. Therefore we can write that,
\[\begin{align}
  & V=\dfrac{dx}{dt} \\
 & \Rightarrow v=-10\pi \sin \left( 2\pi t+\pi \right) \\
\end{align}\]
Substituting the value of time in it will give,
\[\begin{align}
  & v=-10\pi \sin \left( 2\pi \times 1.5+\pi \right) \\
 & \Rightarrow v=-10\pi \sin \left( 4\pi \right) \\
 & \therefore v=0m{{s}^{-1}} \\
\end{align}\]
Now we have to find the acceleration of the object. It can be found by taking the time derivative of the velocity of the body. That is we can write that,
\[a=\dfrac{dv}{dt}\]
Substituting the equation of velocity in it will give,
\[a=-20{{\pi }^{2}}\cos \left( 2\pi t+\pi \right)\]
Substituting the value of time in it can be written as,
\[\begin{align}
  & a=-20{{\pi }^{2}}\cos \left( 2\pi \times 1.5+\pi \right) \\
 & \Rightarrow a=-20{{\pi }^{2}}\cos \left( 4\pi \right) \\
 & \therefore a=-20{{\pi }^{2}} \\
\end{align}\]
From this we can understand that the displacement, velocity and the acceleration of the body at the specified time in the question is found to be as,
\[\begin{align}
  & x=5m \\
 & v=0m{{s}^{-1}} \\
 & a=-20{{\pi }^{2}}m{{s}^{-2}} \\
\end{align}\]

This has been mentioned as option (B).

Note:
The velocity of the body can be found by taking the ratio of the variation of the position of the body to the change in time taken. The velocity is found to be a vector quantity which is having both magnitude as well as direction. The acceleration is also a vector quantity possessing both the magnitude and direction.