Question

The spring extends by x on loading, then energy stored by the spring is (if T is the tension in spring and k spring constant).
A. $\dfrac{{{\text{T}}^{\text{2}}}}{2\text{k}}$
B. $\dfrac{{{\text{T}}^{\text{2}}}}{2{{\text{k}}^{2}}}$
C. $\dfrac{\text{2k}}{{{\text{T}}^{\text{2}}}}$
D. $\dfrac{\text{2}{{\text{T}}^{\text{2}}}}{\text{k}}$

Answer 1

Hint: The answer to this question lies in the knowledge of the energy stored in a string. The formula to find the potential energy stored in the string is very important.

Complete step by step answer:
 According to the given question, we can draw the following diagram for demonstration.

As per the given diagram, we can assume the actual length of the spring as L which is extended by a length of x when a weight of W is added to one end. The equation in the diagram is for the readers to remember this formula in a diagrammatic form. In the given equation, k is the spring constant and x is the increment in the length of the spring. T refers to the tension force created in the spring due to the weight. With the equation,
$\text{k}\cdot \text{x = T}$(given in the diagram)
We can easily calculate the energy stored in the spring.
We know,
Energy in the spring $=\text{ }\dfrac{1}{2}\text{k}{{\text{x}}^{2}}$
$\therefore \text{ E = }\dfrac{1}{2}\cdot \text{k}\cdot {{\left( \dfrac{\text{T}}{\text{k}} \right)}^{2}}\text{ }\left[ \because \,{{\left( \text{k}\cdot \text{x} \right)}^{2}}\text{ = }{{\text{T}}^{2}} \right]$
$\begin{align}
  & =\text{ }\dfrac{1}{2}\cdot \dfrac{{{\text{T}}^{2}}}{\text{k}} \\
 & =\text{ }\dfrac{{{\text{T}}^{2}}}{\text{2k}} \\
\end{align}$
Note: Elastic energy or the potential energy that is stored as a result of the deformation of any elastic object, should always be taken into consideration. By elastic object we mean a string and the phenomenon that we are considering is stretching of a string.
We should know that the spring constant is the measure of the stiffness of the string. From Hooke’s law, we can get an idea about the force that is $\dfrac{{{\text{T}}^{\text{2}}}}{2\text{k}}$ required to find the elastic potential energy.