Question

A simple pendulum has a time period ${{\text{T}}_{\text{1}}}$. The point of suspension is moved upward according to the relation $y=k{{t}^{2}}(k=1m/{{s}^{2}})$, where y is the vertical displacement. The time period now becomes ${{\text{T}}_{\text{2}}}$. The ratio $\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}$ is ,$\left( g=10m/{{s}^{2}} \right)$
A.) $\dfrac{6}{5}$
B.) $\dfrac{5}{6}$
C.) $1$
D.) $\dfrac{4}{5}$

Answer 1

Hint: The time period of a simple pendulum is dependent on its length, the net result acceleration acting on it. It is independent of the mass of the pendulum. If the pendulum is not accelerated, the only acceleration acting on it will be the acceleration due to gravity.

Complete step by step answer:
We know that the time period of a simple pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the net acceleration acting on the body. So, the time period of the simple pendulum can be written as,

$T=2\pi \sqrt{\dfrac{l}{{{a}_{net}}}}$

So, in our problem, the pendulum when it is not accelerated upwards, only the acceleration due to gravity (g) will be acting on it, the time period can be written as,

${{T}_{1}}=2\pi \sqrt{\dfrac{l}{g}}$

In the second case, the same pendulum is moved upwards as a function of time given by, $y=k{{t}^{2}}(k=1m/{{s}^{2}})$. So this movement of the pendulum will lead to the formation of a pseudo force acting on it. We can find this acceleration by differentiating the equation for y.

$\begin{aligned} & y=k{{t}^{2}} \\\ & \dfrac{dy}{dt}=2kt \\\ & \dfrac{{{d}^{2}}y}{d{{t}^{2}}}=2k \\\ \end{aligned}$

So, the acceleration produced due to the motion is, $a=2k$, substituting the value of k we will get, $a=2\text{ m/}{{\text{s}}^{2}}$.

So, the time period of the simple pendulum can be written as,

${{T}_{2}}=2\pi \sqrt{\dfrac{l}{{{a}_{net}}}}=2\pi \sqrt{\dfrac{l}{g+a}}$

The ‘+’ sign indicates that the pendulum is moving upwards.

So the ratio of $\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}$ can be written as,
$\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}=\dfrac{g}{g+a}$

Substituting the values of g and a in the equation above we get,

$\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}=\dfrac{10}{10+2}=\dfrac{5}{6}$
So, the answer to the question option (B)- $\dfrac{5}{6}$.

Note: If the pendulum was accelerated downwards with an acceleration a, the time period of the pendulum will be, $T=2\pi \sqrt{\dfrac{l}{g-a}}$.

For small angle oscillations, the motion of a simple pendulum can be assumed to be a motion similar to a Harmonic oscillator. So the equation can be written as,
$\theta \left( t \right)={{\theta }_{0}}\cos \left( \omega t+\varphi \right)$
$\theta \left( t \right)$ is the angle subtended with the mean position at a time t.
${{\theta }_{0}}$ is the maximum angle that is subtended by the pendulum.
$\omega$ is the angular frequency of the pendulum.
$\varphi$ is the initial phase.