In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

Option 1 - <p>1/2<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>3/4<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>1/3<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>1/4</p>
5 Views|Posted 6 months ago
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1 Answer
A
6 months ago
Correct Option - 2
Detailed Solution:

v = ω√ (A² - x²)
K.E. = ½mv² = ½mω² (A² - x²)
Total Energy T.E = ½mω²A²
At x = A/2
K.E. = ½mω² (A² - (A/2)²) = ½mω²A² (3/4)
K.E./T.E. = (½mω²A² (3/4) / (½mω²A²) = 3/4

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Physics Oscillations 2025

Physics Oscillations 2025

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