Question:
For a particle executing simple harmonic motion, the kinetic energy \(k\) is given by \(k = k_0 \cos^2 \omega t\). The maximum value of potential energy is
For a particle executing simple harmonic motion, the kinetic energy \(k\) is given by \(k = k_0 \cos^2 \omega t\). The maximum value of potential energy is
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In SHM, when kinetic energy is maximum, potential energy is minimum (zero at mean position).
Updated On: Apr 23, 2026
- \(k_0\)
- zero
- \(k_0/2\)
- not obtainable
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
In SHM, kinetic energy and potential energy vary with time and are complementary to each other.
Step 2: Detailed Explanation:
Given \(KE = k_0 \cos^2 \omega t\). When kinetic energy is maximum, \(\cos^2 \omega t = 1\), so \(KE = k_0\).
At this instant, displacement is zero, so potential energy becomes zero.
Step 3: Final Answer:
Thus, maximum value of potential energy is taken as zero.
In SHM, kinetic energy and potential energy vary with time and are complementary to each other.
Step 2: Detailed Explanation:
Given \(KE = k_0 \cos^2 \omega t\). When kinetic energy is maximum, \(\cos^2 \omega t = 1\), so \(KE = k_0\).
At this instant, displacement is zero, so potential energy becomes zero.
Step 3: Final Answer:
Thus, maximum value of potential energy is taken as zero.
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