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. $\frac{2 K ^{{3/2}} A ^{2}}{\sqrt{m}}$
• 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 $(F=K x)$ is:
$P=F V$
$P=(K x) w \sqrt{A^{2}-x^{2}}$
$P=K \sqrt{\frac{K}{m}} \sqrt{A^{2} x^{2}-x^{4}}$
$P$ will be max when $\frac{d p}{d x}=0$
$\Rightarrow A^{2}(2 x)-4 x^{3}=0$
$x=\frac{ A }{\sqrt{2}}$
$P_{\max }=\left.P\right|_{ at x=\frac{A}{\sqrt{2}}}=K \sqrt{\frac{K}{m}} \sqrt{A^{2}\left(\frac{A}{\sqrt{2}}\right)^{2}-\left(\frac{A}{\sqrt{2}}\right)^{4}}$
$P_{\max }=\frac{K^{\frac{3}{2}} A^{2}}{2 \sqrt{m}}$