Question

A simple pendulum of length $l_1$ has frequency $\frac{1}{4} \text{ Hz}$ and another simple pendulum of length $l_2$ has frequency $\frac{1}{3} \text{ Hz}$. Then time period of pendulum of length $(l_1 - l_2)$ is

• A. $5 \text{ s}$
• B. $1 \text{ s}$
• C. $\sqrt{7} \text{ s}$
• D. $\sqrt{12} \text{ s}$

Hint:

For problems involving combinations of lengths, the square of the resulting time period is the combination of the squares of the original time periods. If $L = l_1 + l_2$, then $T = \sqrt{T_1^2 + T_2^2}$.

Answer 1

The Correct Option is C

Step 1: Find individual Time Periods: Frequency ($f$) and Time Period ($T$) are reciprocals ($T = 1/f$).

• For $l_1$: $f_1 = 1/4 \text{ Hz} \implies T_1 = 4 \text{ s}$

• For $l_2$: $f_2 = 1/3 \text{ Hz} \implies T_2 = 3 \text{ s}$

Step 2: Express lengths in terms of Time Periods: Let $k = \frac{g}{4\pi^2}$. Then $l_1 = k T_1^2$ and $l_2 = k T_2^2$. The new length is $L = l_1 - l_2$: $$L = k T_1^2 - k T_2^2 = k(T_1^2 - T_2^2)\lt strong\gt Step 3: Calculate the new Time Period ($T$)\lt /strong\gt T = 2\pi\sqrt{\frac{L}{g}} \implies T^2 = \frac{L}{k}$$ $$T^2 = \frac{k(T_1^2 - T_2^2)}{k} = T_1^2 - T_2^2$$ $$T^2 = 4^2 - 3^2 = 16 - 9 = 7$$ $$T = \sqrt{7} \text{ s}$$ The time period for the pendulum with length $(l_1 - l_2)$ is $\sqrt{7}$ seconds.