Question

Figure shows an ideal spring block system, force constant of spring is k which has been compressed by an amount $x_0$. If $x$ is instantaneous deflection of spring from its natural length, mark the correct option (s).

• A. Instantaneous power developed by spring is $P=kx\sqrt{\frac{k}{m}(x_0^2-x^2)}$
• B. Maximum power of spring is $\frac{k}{2}\sqrt{\frac{k}{m}}{x}_0^2$
• C. Maximum power occurs at $x=\frac{x_0}{\sqrt{2}}$
• D. Maximum power occurs at $x=\frac{x_0}{2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Instantaneous power developed by spring is $P=kx\sqrt{\frac{k}{m}(x_0^2-x^2)}$
B Maximum power of spring is $\frac{k}{2}\sqrt{\frac{k}{m}}{x}_0^2$
C Maximum power occurs at $x=\frac{x_0}{\sqrt{2}}$

$\left(A\right)-\frac{1}{2}k(x^2-x_0^2)=\frac{1}{2}m{v}^2-0$
$\therefore v=\sqrt{\frac{k}{m}(x_0^2-x^2)}$
$\therefore P=F.v=kx\sqrt{\frac{k}{m}(x_0^2-x^2)}$

$\left(C\right)y=x_0^2{x}^2-x^4$
$\frac{dy}{dx}=0\Rightarrow x=\frac{x_0}{\sqrt{2}}$
$\therefore P_{max}$ is at $x=\frac{x_0}{\sqrt{2}}$

$\left(B\right)P=k\sqrt{\frac{k}{m}(x_0^2{x}^2-x^4)}$
$P_{max}=\frac{k}{2}\sqrt{\frac{k}{m}}{x}_0^2$