Question

The magnitude of maximum acceleration is $\pi $ times that of maximum velocity of a simple harmonic oscillator. The time period of the oscillator in seconds is:
(A) 4
(B) 2
(C) 1
(D) $0.5$

Answer 1

Hint: Simple Harmonic Motion arises when we consider the motion of a particle whose acceleration points towards a fixed point O and is proportional to the distance of the particle from O (so the acceleration increases as the distance from the fixed point increases).
As the particle moves away from the fixed point O, since the acceleration is pointing towards O, the particle will slow down and eventually stop (at Q), before returning to O. It will keep going and then again slow down as it reaches P before stopping at P and returning to O once more.
Formula used: The formulae used in the solution are given here.
${v_{\max }} = A\omega $ where ${v_{\max }}$ is the maximum velocity,$\omega $is the angular velocity and $A$ is the amplitude of the motion.
${A_{\max }} = {\omega ^2}A$ where ${A_{\max }}$ is the maximum acceleration.

Complete Step by Step Solution: Simple Harmonic Motion or SHM is defined as a motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards the mean position. The acceleration of a particle executing simple harmonic motion is given by, $a\left( t \right) = - {\omega ^2}x\left( t \right)$. Here, $\omega $ is the angular velocity of the particle.
The angular velocity ($\omega $) refers to how fast an object rotates or revolves relative to another point, i.e. how fast the angular position or orientation of an object changes with time.
We know that, ${v_{\max }} = A\omega $ where ${v_{\max }}$ is the maximum velocity, $\omega $ is the angular velocity and $A$ is the amplitude of the motion.
Maximum acceleration is given by, ${A_{\max }} = {\omega ^2}A$ where ${A_{\max }}$ is the maximum acceleration.
It has been given that the magnitude of maximum acceleration is $\pi $ times that of maximum velocity of a simple harmonic oscillator.
Mathematically, ${A_{\max }} = \pi {v_{\max }}$.
Then, from the above equations, it can be written that, $\omega = \pi $.
The minimum time after which the particle keeps on repeating its motion is known as the time period (or) the shortest time taken to complete one oscillation is also defined as the time period.
 The time period of the oscillator in seconds is $T = \dfrac{{2\pi }}{\pi } = 2s$, since $\omega = \pi $.

So the correct answer is option B.

Note: Two important factors do affect the period of a simple harmonic oscillator.
The period is related to how stiff the system is. A very stiff object has a large force constant ($k$), which causes the system to have a smaller period. For example, you can adjust a diving board’s stiffness—the stiffer it is, the faster it vibrates, and the shorter its period.
Period also depends on the mass of the oscillating system. The more massive the system is, the longer the period. For example, a heavy person on a diving board bounces up and down more slowly than a light one. In fact, the mass m and the force constant k are the only factors that affect the period and frequency of simple harmonic motion.
To derive an equation for the period and the frequency, we must first define and analyse the equations of motion. Note that the force constant is sometimes referred to as the spring constant.