Question

A simple pendulum of length $30$ cm makes $20$ oscillations in $10$ s on a certain planet. Another pendulum makes $40$ oscillations in $10$ s on the same planet. Find the length of the second pendulum.

• A. $10$ cm
• B. $14$ cm
• C. $7.5$ cm
• D. $25$ cm

Hint:

On the same planet, $g$ is constant, so always relate pendulum lengths using $T^2 \propto \ell$.

Answer 1

The Correct Option is C

To solve this problem, we need to understand the relationship between the length of a pendulum and its period of oscillation. The formula for the period of a simple pendulum is given by:

\(T = 2\pi \sqrt{\frac{L}{g}}\)

where \(T\) is the period of one complete oscillation, \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity.

Let's find out the period for each pendulum:

1. The first pendulum completes \(20\) oscillations in \(10\) seconds. Thus, the period \(T_1\) of the first pendulum is: \(T_1 = \frac{10}{20} = 0.5\) seconds.
2. The second pendulum completes \(40\) oscillations in \(10\) seconds. Thus, the period \(T_2\) of the second pendulum is: \(T_2 = \frac{10}{40} = 0.25\) seconds.

Using the formula for the period, we have for the first pendulum:

\(T_1 = 2\pi \sqrt{\frac{L_1}{g}}\)

Substituting the known values,

\(0.5 = 2\pi \sqrt{\frac{30}{g}}\)

Squaring both sides:

\(0.25 = 4\pi^2 \frac{30}{g}\)

Thus,

\(g = 4 \times 0.25 \times 30 \times \pi^2\)

Now, for the second pendulum:

\(T_2 = 2\pi \sqrt{\frac{L_2}{g}}\)

We substitute the period \(T_2 = 0.25\) and solve for \(L_2\):

\(0.25 = 2\pi \sqrt{\frac{L_2}{g}}\)

Squaring both sides:

\(0.0625 = 4\pi^2 \frac{L_2}{g}\)

Substitute \(g\) from the expression obtained earlier:

\(L_2 = \frac{0.0625 \times g}{4\pi^2}\)

Since \(g\) cancels out:

\(L_2 = \frac{0.0625 \times 4 \times 0.25 \times 30}{4} = 7.5 \text{ cm}\)

Thus, the length of the second pendulum is 7.5 cm.