Question

If the displacement, velocity and acceleration of particle in SHM are \[1cm\] , \[1cm/\sec \] and \[1cm/{\sec ^2}\] respectively its time period (in secs) will be:
(A) \[\pi \]
(B) \[0.5\pi \]
(C) \[2\pi \]
(D) \[1.5\pi \]

Answer 1

Hint It is given that the particle undergoes Simple harmonic motion. So, use the equations of SHM to find the angular velocity of the object. And this using angular velocity, you can identify the time period.

Complete Step By Step Solution
In a simple harmonic motion, we consider the particle whose acceleration is pointed towards a fixed point and is proportional to the distance of the particle from the fixed point. When the particle moves away from the fixed point, the particle will slow down and stop at a point since the direction of acceleration is towards the fixed point, and will return back to the fixed point.
So, the general SHM equation for the acceleration of the particle is given as
\[a = {\omega ^2}x\]
Where \[\omega \]is the angular velocity of the object
Now, we know that x= \[1cm\] and a= \[1cm/{\sec ^2}\]
Substituting the value in the equation
\[1 = {\omega ^2} \times 1\]
\[\omega = 1rad/s\]
Now the time period is given as \[T = \dfrac{{2\pi }}{\omega }\] , where T is the time period of the object to undergo one oscillation or one harmonic motion.
Substituting \[\omega \] in the above equation, we get
\[T = \dfrac{{2\pi }}{1}\]
\[T = 2\pi \]
Thus the time period to undergo a simple harmonic motion of given acceleration and given displacement is calculated and found out to be \[2\pi \] .

Hence, Option (c) is the right answer for the given question.

Note
Simple Harmonic motion is the repetitive movement of a body of mass back and forth through an specified equilibrium so that the maximum displacement at one side of its position is equal to the maximum displacement on the other side. Oscillation of pendulum is one of the examples of SHM.