Physics-Simple Harmonic Motion
NEET Physics Simple Harmonic Motion MCQ Question
Type: MCQ-numerical-Medium-Class 11
A simple pendulum with a small angular displacement θ undergoes simple harmonic motion. If the length of the pendulum is 1.5 m and the acceleration due to gravity is 9.8 m/s², what is the angular frequency ω of the pendulum?
A
2.55 rad/s
B
3.29 rad/s
C
2.54 rad/s
D
3.31 rad/s
Correct Answer
Option A
Detailed Explanation
The angular frequency ω of a simple pendulum is given by ω = √(g/L), where g is the acceleration due to gravity and L is the length of the pendulum. Using g = 9.8 m/s² and L = 1.5 m, we find ω = √(9.8/1.5) = 2.55 rad/s.
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