Answer
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Hint: The time period of the pendulum will depend on the acceleration inversely. Now, as the train starts moving, find the pseudo force acting on the pendulum. Next, find the direction of the pseudo acceleration on the pendulum. Thus, we can easily find out the answer.
Formulas used:
$T=2\pi \sqrt{\dfrac{m}{g}}$
Complete answer:
Let us assume the mass of the pendulum as m. If the time period of the pendulum is t, then the relation is,
$T=2\pi \sqrt{\dfrac{l}{g}}$
When the train starts accelerating with uniform acceleration,
The pseudo force will act on the pendulum in the direction opposite to the direction of the train. As a result, we will see the acceleration of the pendulum will also increase. Therefore, the net time period of the pendulum decreases as,
$T\propto \dfrac{1}{\sqrt{({{g}^{2}}+{{a}^{2}})}}$
Where a is the acceleration of the train.
Therefore, the correct option is option b.
Additional information:
The time period of a simple pendulum does not depend on the mass or the initial angular displacement, but depends only on the length of the string and the value of the gravitational field strength. If the simple pendulum is accelerated by putting it in an accelerating car or train, the acceleration of the pendulum will also increase as the net force acting on the simple pendulum will change due to the extra pseudo force. As the acceleration increases, the time taken by the pendulum to reach the maximum points will decrease thus decreasing the time period of the simple pendulum.
Note:
The mass of the car or train or the mass of the pendulum doesn’t affect the time period of the simple pendulum. The net acceleration calculated will be accordingly when the train or car accelerates or decelerates. The pendulum time period depends only on the length of the string and the acceleration of gravity on the pendulum.
Formulas used:
$T=2\pi \sqrt{\dfrac{m}{g}}$
Complete answer:
Let us assume the mass of the pendulum as m. If the time period of the pendulum is t, then the relation is,
$T=2\pi \sqrt{\dfrac{l}{g}}$
When the train starts accelerating with uniform acceleration,
The pseudo force will act on the pendulum in the direction opposite to the direction of the train. As a result, we will see the acceleration of the pendulum will also increase. Therefore, the net time period of the pendulum decreases as,
$T\propto \dfrac{1}{\sqrt{({{g}^{2}}+{{a}^{2}})}}$
Where a is the acceleration of the train.
Therefore, the correct option is option b.
Additional information:
The time period of a simple pendulum does not depend on the mass or the initial angular displacement, but depends only on the length of the string and the value of the gravitational field strength. If the simple pendulum is accelerated by putting it in an accelerating car or train, the acceleration of the pendulum will also increase as the net force acting on the simple pendulum will change due to the extra pseudo force. As the acceleration increases, the time taken by the pendulum to reach the maximum points will decrease thus decreasing the time period of the simple pendulum.
Note:
The mass of the car or train or the mass of the pendulum doesn’t affect the time period of the simple pendulum. The net acceleration calculated will be accordingly when the train or car accelerates or decelerates. The pendulum time period depends only on the length of the string and the acceleration of gravity on the pendulum.
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