Question

A pendulum suspended from the ceiling of the train beats second when the train is at rest. What will be the time period of the pendulum if the train accelerates at $10m{{s}^{-2}}$? Take $g=10m{{s}^{-2}}$.
(a) $\left( 2/\sqrt{2} \right)s$
(b) $2s$
(c) $2/\sqrt{2}s$
(d) None of these

Answer 1

In order to calculate the time period of the pendulum if the train accelerates at $10m{{s}^{-2}}$, we need to use the following formula mentioned below.
When train moves with acceleration a, net effective acceleration of pendulum,
$\sqrt{{{a}^{2}}+{{g}^{2}}}=2\sqrt{10}$

Complete Solution:
Thus,
$T=2\pi \sqrt{l/{{a}_{eff}}}=2\pi \sqrt{L/2\sqrt{1}0}$
$=1/\sqrt{2}\times 1=1/\sqrt{2}$,
Which is none of the options given in the solution.

Thus, the correct answer to this question is option (d).

Additional Information: Simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. Also, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same. Whilst simple harmonic motion is a simplification, it is still a very good approximation. Simple harmonic motion is important in research to model oscillations for example in wind turbines and vibrations in car suspensions.

Note: While solving this question, we should know that frequency of a pendulum is oscillations of it per second and the time period is given by $\dfrac{1}{freq}$. We have to use this formula and put the values of the variables provided in the question carefully in order to get the answer. The acceleration of the pendulum is calculated with different formula which is $\sqrt{{{a}^{2}}+{{g}^{2}}}=2\sqrt{10}$.