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 $P=kx\sqrt{(\frac{k(x_0^2-x^2)}{m})}$
• B. Maximum power of spring is $P=\frac{k{x}_0^2}{2}\sqrt{\frac{k}{m}}$
• 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)

Maximum power occurs at $x=\frac{x_0}{\sqrt{2}}$

Applying conservation of mechanical energy,

$\frac{1}{2}m{v}^2-0=\frac{1}{2}k(x_0^2-x^2)$

$\Rightarrow v=\sqrt{(\frac{k(x_0^2-x^2)}{m})}$

From the definition, Power $P=Fv$

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

To find the maximum power, Let us differentiate the above equation w.r.t $x$ and equate $\frac{dP}{dx}$ to zero.

$2P\frac{dP}{dx}=\frac{2k^2{x}_0^{2}x}{m}-\frac{4k^2{x}^3}{m}$

$\Rightarrow x(x_0^2-2x^2)=0$

$\Rightarrow x=0\text{or}x=\frac{x_0}{\sqrt{2}}$

Substituting this value in $\left(1\right)$ we get,

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

$\Rightarrow P=\frac{k{x}_0^2}{2}\sqrt{\frac{k}{m}}$

Hence, options (a) , (b) and (c) are the correct answers.