NEET Physics Simple Harmonic Motion MCQ Question
A block of mass m is attached to a spring with spring constant k and executes simple harmonic motion with amplitude A. What is the total mechanical energy of the system?
0.5 * k * A^2
m * A^2 / 2
k * A^2 / 4
m * k * A^2
Correct Answer
Detailed Explanation
The total mechanical energy of a simple harmonic oscillator is given by E = 0.5 * k * A^2, where k is the spring constant and A is the amplitude.
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