A particle of mass $m$ is performing linear simple harmonic motion. Its equilibrium is at $x = 0,$ force constant is $K$ and amplitude of $SHM$ is $A$. The maximum power supplied by the restoring force to the particle during $SHM$ will be
A particle of mass $m$ is performing linear simple harmonic motion. Its equilibrium is at $x = 0,$ force constant is $K$ and amplitude of $SHM$ is $A$. The maximum power supplied by the restoring force to the particle during $SHM$ will be
- A
$\frac{{{K^{\frac{3}{2}}}{A^2}}}{{\sqrt m }}$
- B
$\frac{{2{K^{\frac{3}{2}}}{A^2}}}{{\sqrt m }}$
- C
$\frac{{{K^{\frac{3}{2}}}{A^2}}}{{3\sqrt m }}$
- D
$\frac{{{K^{\frac{3}{2}}}{A^2}}}{{2\sqrt m }}$
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A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$, it collies elastically with a rigid wall. After this collision :
$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.
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$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.
$(D)$ the time at which the particle passes througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.
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