Question

A simple pendulum has a time period $T_1$. The point of suspension is now moved upward according to the relation $y=K{{t}^{2}}$, $\left( K=1m{{s}^{-2}} \right)$ where y is the vertical displacement. The time period now becomes $T_2$. The ratio of $\dfrac{{{T}_{1}}^{2}}{{{T}_{2}}^{2}}$ is (Take $g=10m{{s}^{-2}}$ ):
A) $\dfrac{6}{5}$
B) $\dfrac{5}{6}$
C) $\dfrac{1}{2}$
D) $\dfrac{4}{5}$

Answer 1

Hint: The time period of a pendulum depends on the square root of the length of the string of the pendulum from its point of suspension and the square root of the effective downward vertical acceleration. We will use this information to find out the time period ${T_2}$ in terms of ${T_1}$.

Formula used:
$\text{Time period of a pendulum (T) }\propto \text{ }\sqrt{\dfrac{\text{Length of string from point of suspension }\left( L \right)}{\text{Effective downward vertical acceleration }\left( {{g}_{eff}} \right)}}$
$\therefore T\propto \sqrt{\dfrac{L}{{{g}_{eff}}}}$
$\text{acceleration (a) = second derivative of displacement with respect to time (t)}$
$\therefore a=\dfrac{{{d}^{2}}y}{d{{t}^{2}}}$

Complete step-by-step answer:

First we will analyse the given information.
Let the length of the pendulum be L.
Since the pendulum is simple with no other information given, the effective downward acceleration is just the force of gravity = g.
The time period of a pendulum depends on the square root of the length of the string of the pendulum from its point of suspension and the square root of the effective downward vertical acceleration.
$\text{Time period of a pendulum (T) }\propto \text{ }\sqrt{\dfrac{\text{Length of string from point of suspension }\left( L \right)}{\text{Effective downward vertical acceleration }\left( {{g}_{eff}} \right)}}$
$\therefore T\propto \sqrt{\dfrac{L}{{{g}_{eff}}}}$ --(1)
From (1),
${{T}_{1}}=\sqrt{\dfrac{L}{g}}$
$\therefore {{T}_{1}}=\sqrt{\dfrac{L}{10}}$ --- (since given $g=10m{{s}^{-2}}$)
Now, given that the point of suspension is moved upward according to the relation $y=K{{t}^{2}}$, $\left( K=1m{{s}^{-2}} \right)$.
Therefore upward displacement $y=K{{t}^{2}}$, $\left( K=1m{{s}^{-2}} \right)$. --(2)
Now, $\text{acceleration (a) = second derivative of displacement with respect to time (t)}$
$\therefore a=\dfrac{{{d}^{2}}y}{d{{t}^{2}}}$ --(3)
Therefore, using (2) and (3),
$a=2K$
$\therefore a=2m{{s}^{-2}}$
Since this is an acceleration in the upward vertical axis, there will be a pseudo acceleration in the frame of the pendulum in the downward vertical axis along with gravity.
$\therefore {{g}_{eff}}=g+a=10+2=12m{{s}^{-2}}$ --(since given $g=10m{{s}^{-2}}$ )--(4)
The length of the string remains constant L.
So, from (1) and using (4)
New Time period ${{T}_{2}}\propto \sqrt{\dfrac{L}{12}}$
$\therefore \dfrac{{{T}_{1}}^{2}}{{{T}_{2}}^{2}}=\dfrac{\dfrac{L}{10}}{\dfrac{L}{12}}=\dfrac{12}{10}=\dfrac{6}{5}$
Hence, the required ratio is$\dfrac{6}{5}$.
Therefore the correct option is A) $\dfrac{6}{5}$.

Note: The effect of acceleration on the time period of a pendulum is such that an upward acceleration makes the time period smaller while acceleration in the downward direction makes the time period larger. Changing gravity also, has an effect on the time period. A larger gravitational force makes the time period shorter. Thus, a pendulum will have a larger time period on the moon where the gravity is less than that on earth.
A simple experiment to understand the effect of acceleration on the time period would be to tie a small tennis ball to a string and suspend it from your finger. This becomes a pendulum, which can be used by just lifting the ball to some height, keeping the string taut and then letting it go.
Now, go into an elevator. Initially while going up the time period will become fast and become slower midway (when there is no acceleration) and become slowest at the top (when the elevator is decelerating).