A pendulum of length 1 m swings with a small amplitude. What is its approximate

₹0.0

Question: A pendulum of length 1 m swings with a small amplitude. What is its approximate time period? (2023)

Options:

Correct Answer: 2 s

Exam Year: 2023

Solution:

Time period (T) = 2π√(L/g) ≈ 2π√(1/9.8) ≈ 2 s.

A pendulum of length 1 m swings with a small amplitude. What is its approximate time period? (2023)

A pendulum of length 1 m swings with a small amplitude. What is its approximate time period? (2023)

Step 1: Identify the formula for the time period (T) of a pendulum, which is T = 2π√(L/g).
Step 2: Determine the length (L) of the pendulum, which is given as 1 meter.
Step 3: Identify the acceleration due to gravity (g), which is approximately 9.8 m/s².
Step 4: Substitute the values into the formula: T = 2π√(1/9.8).
Step 5: Calculate the value inside the square root: 1/9.8 ≈ 0.10204.
Step 6: Find the square root of 0.10204, which is approximately 0.319.
Step 7: Multiply by 2π: T ≈ 2 * 3.14 * 0.319 ≈ 2.0 seconds.

Pendulum Motion – The time period of a simple pendulum is determined by its length and the acceleration due to gravity.
Small Angle Approximation – The formula for the time period is valid under the assumption of small amplitude oscillations.