Question

At a given place, to increase the number of oscillations made by a simple pendulum in one minute from 72 to 90, the length of the pendulum is to be decreased by

• A. 64%
• B. 36%
• C. 50%
• D. 56%

Hint:

For a simple pendulum, the relationship between frequency ($f$) and length ($L$) is $f \propto 1/\sqrt{L}$. This implies $L \propto 1/f^2$. This proportionality is very useful for ratio-based problems, as you don't need to use the full formula with $g$ and $2\pi$.

Answer 1

The Correct Option is B

The time period ($T$) of a simple pendulum is the time taken for one oscillation. It is related to the length ($L$) by the formula:
$T = 2\pi\sqrt{\frac{L}{g}}$.
The frequency ($f$) is the number of oscillations per unit time, so $f = 1/T$.
From the formula for T, we have $f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$. This shows that $f \propto \frac{1}{\sqrt{L}}$ or $L \propto \frac{1}{f^2}$.
Let the initial state be denoted by subscript 1 and the final state by subscript 2.
Initial frequency $f_1 = 72$ oscillations per minute.
Final frequency $f_2 = 90$ oscillations per minute.
We have the relationship $\frac{L_2}{L_1} = \frac{f_1^2}{f_2^2}$.
$\frac{L_2}{L_1} = \left(\frac{72}{90}\right)^2 = \left(\frac{4 \times 18}{5 \times 18}\right)^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25}$.
So, the new length $L_2$ is $\frac{16}{25}$ of the original length $L_1$. $L_2 = 0.64 L_1$.
The decrease in length is $\Delta L = L_1 - L_2 = L_1 - 0.64 L_1 = 0.36 L_1$.
The percentage decrease in length is given by:
Percentage decrease = $\frac{\Delta L}{L_1} \times 100% = \frac{0.36 L_1}{L_1} \times 100% = 0.36 \times 100% = 36%$.