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Oscillations — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsOscillations2 Marks
Question
How much is KE for displacement equal to half the amplitude?
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Answer

$\because\text{x}=\frac{\text{A}}{2}$ so,

$\text{KE}=\frac{1}{2}\text{m}\omega^2(\text{A}^2-\text{x}^2)$

$\frac{1}{2}\text{m}\omega^2[\text{A}^2-(\frac{\text{A}}{2})^2]$

$=\frac{1}{2}\times\frac{3}{4}[\text{m}\omega^2\text{A}^2]$

$=\frac{3}{4}(\text{KE})_\text{max}$

It is $\frac{3}{4}\text{th}$ of maximum KE.

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