Question

A simple pendulum performs simple harmonic motion about $x=0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x=\dfrac{a}{2}$ will be:
A. $\dfrac{\pi a}{T}$
B. $\dfrac{3\pi^2 a}{T}$
C. $\dfrac{\pi a\sqrt{3}}{T}$
D. $\dfrac{\pi a \sqrt{3}}{2T}$

Answer 1

Recall that a simple harmonic motion entails a to and fro motion of the pendulum about its mean position. We know that the displacement of the pendulum at any instant t is given by the periodic equation $x = a\;sin(\omega t)$. The rate of change of this equation will give the speed of the pendulum at any instant of time t. Rewrite and rearrange the equation that you get to suit the demands of the problem. Following this, a straightforward values substitution should give you the required speed of the pendulum.

Formula Used:
Speed of pendulum oscillations: $v = \omega\sqrt{a^2-x^2}$

Complete step by step answer:
We are given that this simple pendulum carries out a simple harmonic motion (SHM).
SHM is a type of periodic motion where the force acting on the oscillating object is directly proportional to the magnitude of the object’s displacement and is always directed towards the object’s equilibrium position. This is given as:
$\vec{F}_{restoring} \propto -\vec{x}$, where the negative sign indicates that the restoring force is directed towards the pendulum’s equilibrium position.

For any SHM, the displacement of the pendulum at any instant of time t is given by:
$x = a\;sin(\omega t)$, where a is the amplitude of oscillation (which is the maximum displacement of the pendulum from mean position) and $\omega$ is the angular frequency.
The speed of the pendulum oscillations at any instant of time t is given by taking the derivative of the above equation, i.e.,
$v = \dfrac{dx}{dt}=\dfrac{d}{dt} (a\;sin(\omega t)) = a\omega\;cos\omega t$
This can be re-written as: $v = a\omega\sqrt{1 – sin^2(\omega t)} = \omega\sqrt{a^2 – a^2 sin^2(\omega t)}$
Substituting $x = a\;sin(\omega t)$ in the above equation, we get:
$v = \omega\sqrt{a^2-x^2}$
Thus, for any body executing simple harmonic motion about its position, the speed of oscillations is given by :
$v = \omega\sqrt{a^2-x^2}$, where$\omega =2\pi f = \dfrac{2\pi}{T}$, where a is the amplitude of oscillation, $\omega$ is the angular frequency of oscillation, f is the frequency of oscillation and T is the time period, and x is the displacement.
At $x = \dfrac{a}{2}$:
$v = \dfrac{2\pi}{T}\sqrt{a^2 - \left(\dfrac{a}{2}\right)^2} = \dfrac{2\pi}{T}\sqrt{a^2 - \left(\dfrac{a^2}{4}\right)} = \dfrac{2\pi}{T}\sqrt{\dfrac{4a^2 – a^2}{4}} =\dfrac{2\pi}{T}\sqrt{\dfrac{3a^2}{4}} = \dfrac{2\pi}{T}\dfrac{a}{2}\sqrt{3}$
$\Rightarrow v = \dfrac{\pi a\sqrt{3}}{T}$
Therefore, the correct choice would be C. $\dfrac{\pi a\sqrt{3}}{T}$.

Note:
Do not get confused between periodic, oscillatory and harmonic motion. The motion which is repeated at regular intervals is called periodic motion. The periodic motion in which a body moves to and fro about its fixed position is called oscillatory motion. The oscillatory motion in which the restoring force or acceleration is always directed towards the equilibrium position is called harmonic motion. Note that all oscillatory (vibratory) motion are periodic but all periodic motion need not be oscillatory. For example, the revolution of the earth around the sun is periodic but not oscillatory. Also, all periodic motions need not be simple harmonic.
Remember that the mean position of the body, where no net force acts on it is called the equilibrium position, and the point at which maximum restoring force is acting on the body is called the extreme position.