Question

A particle of mass $m$ is performing the 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. $\frac{2K^{3/2}{A}^2}{\sqrt{m}}$
• C. $\frac{K^{\frac{3}{2}}{A}^2}{3\sqrt{m}}$
• D. $\frac{K^{\frac{3}{2}}{A}^2}{2\sqrt{m}}$

Answer 1

Answer: (D)

Power supplied by the restoring force $(F=Kx)$ is:
$P=FV$
$P=(Kx)w\sqrt{A^2-x^2}$
$P=K\sqrt{\frac{K}{m}}\sqrt{A^2{x}^2-x^4}$
$P$ will be max when $\frac{dp}{dx}=0$
$\Rightarrow A^2(2x)-4x^3=0$
$x=\frac{A}{\sqrt{2}}$
$P_{\max }={P|}_{atx=\frac{A}{\sqrt{2}}}=K\sqrt{\frac{K}{m}}\sqrt{A^2{\left(\frac{A}{\sqrt{2}}\right)}^2-{\left(\frac{A}{\sqrt{2}}\right)}^4}$
$P_{\max }=\frac{K^{\frac{3}{2}}{A}^2}{2\sqrt{m}}$