A pendulum suspended from the ceiling of a train has a time period t when the train is at rest. When the train is accelerating with a uniform acceleration, the time period will
a)Increase
b)Decrease
c)Remains constant
d)None
Answer
655.5k+ views
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.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

