Two spring $P$ and $Q$ of force constants $k_{p}$ and $k_{Q}\left( k_{Q} = \frac{k_{p}}{2} \right)$ are stretched by applying forces of equal magnitude if the energy stored in $Q$ is $E$, then the energy stored in $P$ is

Question

Dr. Abubacker · Accepted Answer

(d) Here, $k_{Q} = \frac{k_{p}}{2}$
According to Hooke’s law
$	herefore F_{p} = - k_{p}x_{p}$
$F_{Q} = - k_{Q}x_{Q} \Rightarrow \frac{F_{p}}{F_{Q}} = \frac{k_{p}}{k_{Q}}\ \frac{x_{p}}{x_{Q}}$
$F_{p} = F_{Q}$ [Given]
$	herefore\frac{x_{p}}{x_{Q}} = \frac{k_{Q}}{k_{p}}$ ….(i)
Energy stored in a spring is $U = \frac{1}{2}kx^{2}$
$	herefore\frac{U_{P}}{U_{Q}} = \frac{k_{p}x_{p}^{2}}{k_{Q}x_{Q}^{2}} = \frac{k_{p}}{k_{Q}} 	imes \frac{k_{Q}^{2}}{k_{p}^{2}} = \frac{1}{2}\ \ \ \left\lbrack \because k_{Q} = \frac{k_{p}}{2} ightbrack$
$\Rightarrow U_{p} = \frac{U_{Q}}{2} = \frac{E}{2}\ \ \lbrack\because U_{Q} = Ebrack$

Q. Two spring $P$ and $Q$ of force constants $k_{p}$ and $k_{Q}\left( k_{Q} = \frac{k_{p}}{2} \right)$ are stretched by applying forces of equal magnitude if the energy stored in $Q$ is $E$, then the energy stored in $P$ is

Explanation

Reviewed by Dr. Abubacker

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