Potential energy of spring: When a spring is compressed or elongated by a force F and the elongation or compression is x, then F x.

Or, F = kx [where k is the force constant of spring] ——-(i)

If F_{e} is the elastic force or restoring force applied by spring, then F_{e} = – kx [F_{e} is always towards the mean position].

Now the spring is further stretched through a distance dx, therefore the work done by the spring for further stretching is dw = -F_{e}.dx = F_{e}dxcos180^{0} = -F_{e}dx = -kxdx [using equation (i)]

Therefore the work done to stretched the spring through a distance x from its normal position(x = 0) is

W = dW = –kxdx = -k[] = – kx^{2} = potential energy of spring.

If spring is elongated from x_{i} to x_{f}, then work done by spring is w = – k( – ).

Conservation of energy of spring: When a spring is at unstretched condition, the potential energy of the spring is zero. Now F force is applied and the elongation of it is x and the potential energy of spring is kx^{2}. At this time spring is at extreme position. Now the spring moves to its normal position due to elastic property, and at that condition total energy is kinetic energy i.e. kx^{2}. When the spring is in extreme compressed position again total energy is converted to potential i.e. kx^{2} and this process continues. So energy is conserved in vibrating spring.

Force constant of spring calculation:

1. Spring in series: Two massless springs of force constant k_{1} and k_{2} are connected in series as shown in figure. Due to applied force F on the body the body is displaced by x. As the springs are massless, so the force acts on each spring is also F. If x_{1} and x_{2} are the elongation of two springs of force constant k_{1} and k_{2} respectively, then

F = k_{1}x_{1} and F = k_{2}x_{2}.

But x_{1} + x_{2} = x. So, + = x

If k is the equivalent force constant [replacing two springs by a single spring keeping force and elongation same] then, + =

Therefore, in series connection of spring = + .

2. Spring in parallel: (i) Two massless springs of force constant k_{1} and k_{2} are connected in parallel as shown in figure. Due to applied force F, the body is displaced by x.

Therefore, the elongation of each spring is also x.

If F_{1} and F_{2} are the force applied on two springs of force constant k_{1} and k_{2} respectively, then F_{1} = k_{1}x and F_{2} = k_{2}x.

Here F = F_{1} + F_{2} = k_{1}x + k_{2}x = (k_{1} + k_{2})x

If k is the equivalent force constant [replacing two spring by a single spring keeping force and elongation same] then, kx = (k_{1} + k_{2})x. Therefore, in parallel connection of spring k = k_{1} + k_{2}.

(ii) In this figure when F force is applied on body towards right to displaced it by x, then spring k_{1} is expanded by x where F_{1} is the spring force produced. Similarly, spring k_{2} is compressed by x where F_{2} is the spring force.

Therefore, F = F_{1} + F_{2} = k_{1}x + k_{2}x = (k_{1} + k_{2})x

If k is the equivalent force constant [replacing two spring by a single spring keeping force and elongation same] then, kx = (k_{1} + k_{2})x

Therefore, the equivalent force constant k = k_{1} + k_{2}.