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Question: The length of a simple pendulum for a given time period is \(l\). If the pendulum is taken to a plac...

The length of a simple pendulum for a given time period is ll. If the pendulum is taken to a place where acceleration due to gravity is doubled, the for the time period to remain the same its length should be:
A. 2ll
B. ll
C. l2\dfrac{l}{2}
D. l4\dfrac{l}{4}

Explanation

Solution

Hint: Use the formula of time period of simple pendulum because the relation between time period, length and acceleration due to gravity are contained in it given by
T = 2πlg{\text{T = 2}}\pi \sqrt {\dfrac{l}{g}}
When the gravity is doubled then the time period has decreased so according to the above relation, the length has to increase in order to maintain the same time period.

Complete step-by-step answer:
According to question we know that,
As, we change the acceleration due to gravity then the force acting on it changes and hence the time period also changes.
So, in order to remain time period constant we have to change the length also according to the given condition.
The time of a simple pendulum depends on the length and acceleration due to gravity. The relation is given as follow: -
T = 2πlg{\text{T = }}2\pi \sqrt {\dfrac{l}{g}}
Where
ll= length of pendulum
g= acceleration due to gravity
Now, we have to convert this relation a little bit to find variation of length with the acceleration due to gravity and time period.
So, on squaring both sides we get
T2 = 4π2lg{{\text{T}}^{\text{2}}}{\text{ = }}4{\pi ^2}\dfrac{l}{g}
Solving for ll we have
l=T2g4π2(1)l = \dfrac{{{{\text{T}}^{\text{2}}}g}}{{4{\pi ^2}}} \cdot \cdot \cdot \cdot \cdot \left( 1 \right)
Let the new length of the pendulum be l{l^{‘}}
Also, let new acceleration due to gravity be g’=2g
New equation of time period given by
l=T2g4π2{l^{'}} = \dfrac{{{T^2}g'}}{{4{\pi ^2}}}
Now on putting values in equation of new length and acceleration due to gravity we have
l=T2(2g)4π2{l^{'}} = \dfrac{{{T^2}\left( {2g} \right)}}{{4{\pi ^2}}}
l=2T2g4π2(2){l^{'}} = 2\dfrac{{{T^2}g}}{{4{\pi ^2}}} \cdot \cdot \cdot \cdot \cdot \left( 2 \right)
From equation (1) and (2) we have
l=2l{l^{'}} = 2l
\therefore Option (A) is correct.

Note: By changing the acceleration due to gravity changes the value of force acting on the pendulum as a result its time period will change. So to counter this we change the length of the pendulum because it changes the tension force. Also do not forget to square the formula for ease in calculation. We can only vary the time period of the pendulum by changing the length of the pendulum and acceleration due to gravity.