Question

A body is in simple harmonic motion with a time period half second (T=0.5s) and amplitude one cm (A= 1cm). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.
A. 4 cm /s
B. 6 cm /s
C. 12 cm /s
D. 16 cm /s

Answer 1

Hint: In order to solve these types of questions we need a good understanding of what simple harmonic motion is with its correlated terms and its mathematical general equation. Afterward, substitute the given values in the general equation of SHM and thus find the missing physical quantities. Finally, substitute the found required values in the formula of average velocity and solve for the required solution.

Formulas used:
Equation of SHM, $x = A\sin \left( {\omega t + \delta } \right)$
Angular velocity, $\omega = \dfrac{{2\pi }}{T}$
Average velocity, ${v_{avg}} = \dfrac{{{\text{Displacement}}}}{{{\text{time taken}}}} = \dfrac{x}{t}$

Complete step-by-step answer:
Simple harmonic motion (SHM) is a special type of oscillation i.e. to and fro motion in which the particle oscillates on a straight line, the acceleration of the particle is always directed towards a fixed point on the line and its magnitude is proportional to the displacement of the particle from this point. This fixed point is called the center of oscillation.
Equation of a body undergoing SHM is given by:
$x = A\sin \left( {\omega t + \delta } \right)$
where $A$ is the amplitude of the SHM,
$\omega t$is the angle that increases with time,
$\delta $ is the phase difference

Time taken by an oscillating body to complete one whole oscillation is called the time period, T of that oscillating system.
Amplitude, A of an oscillating body is the maximum displacement of that body from the center of oscillation.
Average velocity is nothing but the ratio of the displacement of a body upon the time taken.
Now, we know that
$x = A\sin \left( {\omega t} \right){\text{ where }}\omega = \dfrac{{2\pi }}{T}{\text{ }} \ldots \ldots \ldots \ldots \left( 1 \right){\text{ }}\left[ {\because {\text{here }}\delta = 0} \right]$
If the particle starts at the mean position at t = 0, also it is given that $x = \dfrac{A}{2}$
Substituting the value of x in equation (1), we get:
$\eqalign{
  & \dfrac{A}{2} = A\sin \omega t \cr
  & \Rightarrow \sin \omega t = \dfrac{1}{2} \cr
  & \Rightarrow \omega t = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \cr} $
But $\omega = \dfrac{{2\pi }}{T}$
$\eqalign{
  & \Rightarrow \dfrac{{2\pi }}{T}t = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \cr
  & \Rightarrow \dfrac{{2\pi }}{T}t = \dfrac{\pi }{6} \cr
  & \Rightarrow t = \dfrac{1}{{2 \times 2 \times 6}}{\text{seconds }}\left[ {\because T = \dfrac{1}{2}{\text{s (given)}}} \right] \cr
  & \Rightarrow t = \dfrac{1}{{24}}{\text{seconds}} \cr} $
Now, the required total displacement is$\dfrac{A}{2}$
Hence average velocity of the oscillating body is:
$\eqalign{
  & {v_{avg}} = \dfrac{{{\text{Displacement}}}}{{{\text{time taken}}}} = \dfrac{x}{t} \cr
  & \Rightarrow {v_{avg}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{{24}}}} = \dfrac{1}{2} \times \dfrac{{24}}{1}{\text{ }}\left[ {\because A = 1{\text{ }}cm{\text{ }}\left( {given} \right)} \right] \cr
  & \Rightarrow {v_{avg}} = 12{\text{ m/s}} \cr} $
Therefore, the correct answer is C. i.e., 12m/s.

Note: Students commonly make the mistake of confusing between the time period T, and the time taken to undergo certain displacement, t. Both T and t variables clearly represent two completely different quantities. Thus must not be confused with. Also, avoid silly calculation errors by rechecking your calculations.