Question

The ratio of the frequencies of two simple pendulums is 4 : 3 at the same place. The ratio of their respective lengths is ______.

• A. 3 : 4
• B. 4 : 3
• C. 9 : 16
• D. 16 : 9

Hint:

Frequency is inversely proportional to the square root of length. So, if a pendulum swings FASTER (higher frequency), it must be SHORTER. Pendulum 1 is faster (4:3), so it must be shorter (9:16).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the ratio of frequencies of two pendulums swinging under the same gravity. We must calculate the inverse ratio involving their physical lengths.

Step 2: Detailed Explanation:
The frequency ($f$) of a simple pendulum is the inverse of its time period ($T$).
Since $T = 2\pi \sqrt{\frac{L}{g}}$, the frequency is given by:
$f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$
Because both pendulums are at the "same place", the acceleration due to gravity ($g$) is completely constant for both.
Therefore, the frequency is strictly inversely proportional to the square root of the pendulum's length:
$f \propto \frac{1}{\sqrt{L}}$
We can set up a ratio for the two pendulums:
$\frac{f_1}{f_2} = \frac{\sqrt{L_2}}{\sqrt{L_1}} = \sqrt{\frac{L_2}{L_1}}$
We are given the frequency ratio $\frac{f_1}{f_2} = \frac{4}{3}$.
Substitute this into the equation:
$\frac{4}{3} = \sqrt{\frac{L_2}{L_1}}$
To remove the square root, square both sides of the equation:
$\left(\frac{4}{3}\right)^2 = \frac{L_2}{L_1}$
$\frac{16}{9} = \frac{L_2}{L_1}$
The question asks for the ratio of their respective lengths, which means $L_1 : L_2$.
Simply invert the fraction:
$\frac{L_1}{L_2} = \frac{9}{16}$

Step 3: Final Answer:
The ratio of their lengths is 9 : 16, matching option (c).