Question

A simple pendulum suspended from the ceiling of a stationary lift has period \[{T_0}\]. When the lift descends at steady speed, the period is \[{T_1}\], and when it descends with constant downward acceleration, the period is \[{T_2}\]. Which of the following is true?
A. \[{T_0} = {T_1} = {T_2}\]
B. \[{T_0} = {T_1} < {T_2}\]
C. \[{T_0} = {T_1} > {T_2}\]
D. \[{T_0} < {T_1} < {T_2}\]

Answer 1

Hint:Use the expression for Newton’s second law of motion. Also use the formula for the time period of the simple pendulum. This formula gives the relation between the time period of a simple pendulum, length of the simple pendulum and acceleration due to gravity. Hence, calculate the value of acceleration due to gravity in each case and derive the formula for time period in each case and compare all the three formulae for time period to determine relation between them.

Formulae used:
The expression for Newton’s second law of motion is
\[{F_{net}} = ma\] …… (1)
Here, \[{F_{net}}\] is net force acting on the object, \[m\] is mass of the object and \[a\] is acceleration of the object.
The time period \[T\] of a simple pendulum is
\[T = 2\pi \sqrt {\dfrac{L}{g}} \] …… (2)
Here, \[L\] is the length of the simple pendulum and \[g\] is acceleration due to gravity.

Complete step by step answer:
We have given that the time period of the simple pendulum suspended from the ceiling of the lift is \[{T_0}\]. The time period of the same simple pendulum when the lift is moving in the downward direction with a constant speed is \[{T_1}\] and time period of the same simple pendulum when the lift is moving downward with a constant acceleration is \[{T_2}\].When the lift is steady, the weight of the simple pendulum in the downward direction is balanced by tension \[T'\] in the string acting in the upward direction.
\[T' = mg\]
\[ \Rightarrow g = \dfrac{{T'}}{m}\]
Hence, according to equation (2), the time period \[{T_0}\] of the simple pendulum becomes
\[{T_0} = 2\pi \sqrt {\dfrac{L}{{\dfrac{{T'}}{m}}}} \]
\[ \Rightarrow {T_0} = 2\pi \sqrt {\dfrac{{mL}}{{T'}}} \] …… (3)

When the lift is moving in the downward direction with a constant speed then also the tension in the pendulum is balanced by the weight of the pendulum as the acceleration of the simple pendulum is zero as it is moving with the constant speed. The only acceleration acting on the pendulum is acceleration due to gravity.
Hence, according to equation (2), the time period \[{T_1}\] of the simple pendulum becomes
\[ \Rightarrow {T_1} = 2\pi \sqrt {\dfrac{{mL}}{{T'}}} \] …… (4)

Now consider the situation when the simple pendulum is moving downward with a constant acceleration \[a\].Let us apply Newton’s second law of motion to this pendulum
\[T' - mg = - ma\]
\[ \Rightarrow T' + ma = mg\]
\[ \Rightarrow g = \dfrac{{T' + ma}}{m}\]
From the above equation, we can see that the net acceleration of the simple pendulum is \[\left( {g - a} \right)\].
Hence, according to equation (2), the time period \[{T_2}\] of the simple pendulum becomes
\[ \Rightarrow {T_2} = 2\pi \sqrt {\dfrac{{mL}}{{T' + ma}}} \] …… (5)
Let us now compare the time periods \[{T_0}\], \[{T_1}\] and \[{T_2}\] of the simple pendulum, we get
\[\therefore {T_0} = {T_1} < {T_2}\]

Hence, the correct option is B.

Note:The students should keep in mind that in the second case the velocity of the lift is constant, so the acceleration of the lift is zero which is the same case as that when the lift was at rest. But in the third case the lift has an additional downward acceleration along with the acceleration due to gravity. The students should not forget to consider this acceleration.