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Q.
Two springs of spring constant $1500 \,N / m$ and $3000 \,N / m$ respectively are stretched with the same force. They will have the potential energies in the ratio of :
AFMCAFMC 2000
Solution:
When a body oscillates in simple harmonic. motion, it is acted upon by a restoring force which tends to bring it in the equilibrium position. Due to this force there is potential energy in the body.
Restoring force acting on body is
$F=$ mass $\times$ acceleration
$=m \omega^{2} x$
where, $\omega$ is angular frequency and $x$ is displacement.
The work done in displacing the body appears as potential energy $U$.
$U=W=\int\limits_{0}^{\alpha} m \omega^{2} x d x=m \omega^{2} \frac{x^{2}}{2}$
Here, $m \omega^{2}=k=$ spring constant.
$\therefore U_{1}=\frac{1}{2} k_{1} x^{2}, $
$U_{2}=\frac{1}{2} k_{2} x^{2}$
$\therefore \frac{U_{1}}{U_{2}}=\frac{1500}{3000}=\frac{1}{2}$
$\therefore U_{1}: U_{2}=1: 2$
