JEE MAIN - Physics (2023 - 11th April Morning Shift - No. 9)
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement $$(x)$$ starting from mean position to extreme position (A) is given by
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Explanation
For a particle executing SHM
$$ \mathrm{KE}=\frac{1}{2} \mathrm{m} \omega^2\left(\mathrm{A}^2-\mathrm{x}^2\right) $$
When $\mathrm{x}=0, $ KE is maximum & when $\mathrm{x}=\mathrm{A}, $ KE is zero and KE vs x graph is parabola.
$$ \mathrm{KE}=\frac{1}{2} \mathrm{m} \omega^2\left(\mathrm{A}^2-\mathrm{x}^2\right) $$
When $\mathrm{x}=0, $ KE is maximum & when $\mathrm{x}=\mathrm{A}, $ KE is zero and KE vs x graph is parabola.
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