Question

To increase the period of a pendulum by 10 percent, the length should be increase by (in percentage):
A. 21
B. 11
C. 10.5
D. 10

Answer 1

Hint: Recall that the time period of a pendulum is directly proportional to the length of the pendulum and inversely proportional to acceleration due to gravity under root. In such a case, assuming that the period increases by ten percent, proportionally determine the change in length under that condition by equating the proportionalities as ratios to arrive at the appropriate result.

Formula used: Time period of a simple pendulum $T = 2\pi\sqrt{\dfrac{L}{g}}$

Complete step by step answer:
We know that a simple pendulum is a point mass that is suspended from a string that is attached to a fixed pivot and executes a simple harmonic motion (SHM).
SHM is a type of periodic motion where the force acting on the oscillating object is directly proportional to the magnitude of the object’s displacement and is always directed towards the object’s equilibrium position. The time period of the pendulum is the time taken by the pendulum to finish one full oscillation from its mean position.
Time period of a simple pendulum is thus given as:
$T = 2\pi\sqrt{\dfrac{L}{g}}$, where L is the length of the string of the pendulum and g is the acceleration due to gravity.
From the above relation, we have that $T \propto \sqrt{L}$
Now, for the first case, let the time period be T and the length of the pendulum be L.
$T \propto \sqrt{L}$
For the second case we have the time period :
$T^{\prime} \propto \sqrt{L^{\prime}} $
$\Rightarrow T + \dfrac{10}{100}T \propto \sqrt{L^{\prime}}$
$\Rightarrow 1.1T \propto \sqrt{L^{\prime}}$
Dividing the two equations from each case, we have:
$\dfrac{T}{1.1.T} = \dfrac{\sqrt{L}}{\sqrt{L^{\prime}}}$
Squaring both the sides we get:
$\dfrac{T^2}{1.21T^2} = \dfrac{L}{L^{\prime}}$
$\Rightarrow L^{\prime} = 1.21L$
Therefore, the percentage increase in length is:
$\dfrac{L^{\prime} - L}{L}\times 100 = \dfrac{1.21L-L}{L} \times 100 = \dfrac{0.21L}{L} \times 100 = 21\%$

So, the correct answer is “Option A”.

Note: Remember to obtain the resultant change in length as a percentage since that is what is ultimately asked in the question.
Also, do not forget that we make a number of assumptions while approximating the time period of the simple pendulum to the relation that we have used in the problem. These are:
1. There is negligible friction between the air and the pendulum
2. The string of the pendulum is massless, and does not bend or compress
3. The pendulum executes SHM in a perfect plane
4. Gravity remains constant throughout