Question

Savitha notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take $\pi^2 = 9.8$, and $g = 9.8$ m/s$^2$)

• A. 2 m
• B. 1 m
• C. 0.75 m
• D. 1.5 m

Hint:

A pendulum with a time period of exactly 2 seconds is called a "Seconds Pendulum." Its length is always approximately 1 meter on Earth.

Answer 1

The Correct Option is B

Step 1: Understanding the Topic:
This problem is from the chapter "Oscillations." It involves a simple pendulum, a classic harmonic oscillator. The motion of a simple pendulum is periodic, and its time period depends solely on the length of the string and the local acceleration due to gravity, provided the angle of swing is small.
Step 2: Key Formulas and Approach:

Time Period ($T$) = $\text{Total time taken } / \text{Number of oscillations}$.
Pendulum Formula: $T = 2\pi \sqrt{\frac{L}{g}}$.
Squaring gives: $T^2 = \frac{4\pi^2 L}{g}$.

Step 3: Detailed Explanation:

Calculate the Time Period ($T$): Savitha observed 30 oscillations in 60 seconds. The period is the time required for exactly one oscillation. \[ T = \frac{60 \text{ s}}{30} = 2 \text{ seconds} \]
Identify constants: We are told to take $\pi^2 = 9.8$ and $g = 9.8 \text{ m/s}^2$. This means $\pi^2 \approx g$, which simplifies the algebra significantly.
Rearrange for length ($L$): \[ L = \frac{T^2 \cdot g}{4 \cdot \pi^2} \]
Substitute the values: \[ L = \frac{(2)^2 \times 9.8}{4 \times 9.8} \]
Notice that 9.8 cancels out from the numerator and denominator: \[ L = \frac{4}{4} = 1 \text{ meter} \]
This specific type of pendulum, which has a period of 2 seconds, is historically significant and is known as a "Seconds Pendulum."
Step 4: Final Answer:
The calculated length of the simple pendulum is 1 m.