Q. Two springs of spring constant $1500\, Nm^{ - 1} \, and \, 3000\, Nm ^{-1}$ respectively are stretched with the same force. They will have potential energy in ratio
Solution:
The work done in pulling the string is stored as potential energy in the spring
$U =\frac{1}{2} k x^{2}\,\,\,...(i)$
where $k$ is spring constant and $x$ is distance through which it is pulled.
Also in SHM
Force $\propto$ displacement
$F= kx \,\,\,...(ii)$
where $k$ is spring constant
Putting $x=\frac{F}{k}$ in Eq. (i), we get
$U =\frac{1}{2} k\left(\frac{F}{2}\right)^{2}=\frac{F^{2}}{2 k}$
$\therefore \frac{U_{1}}{U_{2}}=\frac{k_{2}}{k_{1}}=\frac{3000}{1500}=\frac{2}{1}$
$\therefore U_{1}: U_{2}=2: 1$
