Question

A particle is undergoing simple harmonic motion on a straight line. If its start moving from extreme position, then distance travelled by the particle in time $\dfrac{5T}{6}$ will be (where A is amplitude and T is time period)
$\text{A}\text{. }\dfrac{\left( 5-\sqrt{3} \right)A}{2}$
B. zero
$\text{C}\text{. }\dfrac{A}{2}$
$\text{D}\text{. }\dfrac{7A}{2}$

Answer 1

Hint: Use the equation for the position of a particle undergoing a simple harmonic motion, when it starts its motion from one of the extreme. Calculate the position of the particle at a given time. Then measure the total distance covered by the particle in this time.
Formula used:
$x=A\cos \omega t$
$T=\dfrac{2\pi }{\omega }$

Complete answer:
It is given that a particle is undergoing a simple harmonic motion on a straight line. The amplitude of the motion is given to be A and the time period of the motion is given as T.
When a particle starts a simple harmonic motion from the extreme position, the equation for its position with respect to time is given as $x=A\cos \omega t$ ….. (i),
where $\omega $ is the angular frequency of the oscillations.
This means that at time t=0, the position of the particle is A.
The relation between time period (T) and angular frequency ($\omega $) is given as $T=\dfrac{2\pi }{\omega }$.
Let the find the position of the particle at time $t=\dfrac{5T}{6}=\dfrac{5}{6}\left( \dfrac{2\pi }{\omega } \right)=\dfrac{5\pi }{3\omega }$.
Substitute the value of t in equation (i).
$\Rightarrow x=A\cos \omega \left( \dfrac{5\pi }{3\omega } \right)$
$\Rightarrow x=A\cos \left( \dfrac{5\pi }{3} \right)$
$\Rightarrow x=A\cos \left( 2\pi -\dfrac{\pi }{3} \right)$
$\Rightarrow x=A\cos \left( \dfrac{\pi }{3} \right)=\dfrac{A}{2}$
This means that the position of the particle at time $\dfrac{5T}{6}$ is $\dfrac{A}{2}$.
$\dfrac{5T}{6}$can be written as $\dfrac{T}{2}+\dfrac{T}{4}+\dfrac{T}{12}$.
In time $\dfrac{T}{2}$, the particle will oscillate from one extreme to the other. Then in next $\dfrac{T}{4}$ seconds, it will come at the mean position. In another $\dfrac{T}{12}$ seconds, it will be at $\dfrac{A}{2}$.
The distance from one extreme to another is 2A and the distance between the extreme and the mean position is A.
Therefore, the total distance moved by the particle is $2A+A+\dfrac{A}{2}=\dfrac{7A}{2}$.

So, the correct answer is “Option D”.

Note:
One of the mistake that a student may perform is that he or she may write the total distance travelled by the particle equal to $\dfrac{A}{2}$, after finding the position of the particle at time t = $\dfrac{5T}{6}$. Note that $\dfrac{A}{2}$ is the position of the particle with respect to the mean position and not the distance covered.