JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 12)
The maximum potential energy of a block executing simple harmonic motion is $$25 \mathrm{~J}$$. A is amplitude of oscillation. At $$\mathrm{A / 2}$$, the kinetic energy of the block is
9.75 J
37.5 J
18.75 J
12.5 J
Explanation
$\mathrm{U}_{\max }=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}=25 \mathrm{~J}$
$\mathrm{KE}$ at $\frac{\mathrm{A}}{2}=\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-\frac{A^{2}}{4}\right)$
$=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{3 \mathrm{~A}^{2}}{4}=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$
$\mathrm{KE}=\frac{3}{4} \times 25=18.75 \mathrm{~J}$
$\mathrm{KE}$ at $\frac{\mathrm{A}}{2}=\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-\frac{A^{2}}{4}\right)$
$=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{3 \mathrm{~A}^{2}}{4}=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$
$\mathrm{KE}=\frac{3}{4} \times 25=18.75 \mathrm{~J}$
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