A particle executing a simple harmonic motion | vedclass

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(A) Let the initial amplitude be $A$. The total energy of the particle in simple harmonic motion is $E = \frac{1}{2} k A^2$.At the mean position,the p

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its amplitude will change and become equal to $\sqrt{2}$ times its previous amplitude

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(A) Let the initial amplitude be $A$. The total energy of the particle in simple harmonic motion is $E = \frac{1}{2} k A^2$.At the mean position,the potential energy is $0$ and the kinetic energy is $E$. Thus,the additional energy received is $E$.At the extreme position,the initial energy is $E$ (all potential). After receiving the additional energy $E$,the new total energy becomes $E' = E + E = 2E$.Since $E = \frac{1}{2} k A^2$,we have $E' = \frac{1}{2} k (A')^2 = 2 	imes (\frac{1}{2} k A^2)$.This implies $(A')^2 = 2A^2$,so the new amplitude $A' = \sqrt{2} A$.The period of simple harmonic motion depends only on the mass and the force constant $(T = 2\pi \sqrt{m/k})$,so it remains unchanged.

$A$ particle executing a simple harmonic motion of period $2 \ s$. When it is at its extreme displacement from its mean position,it receives an additional energy equal to what it had in its mean position. Due to this,in its subsequent motion,

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