If a spring extends by `x` on loading, then the energy stored by the spring is (if T is tension in the spring and k is spring constant)

Options

A. `(T^(2))/(2x)`
B. `(T^(2))/(2K)`
C. `(2x)/(T^(2))`
D. `(2T^(2))/(2k)`

Explanation

To find the energy stored in a spring when it extends by a distance \( x \), we can use the formula for elastic potential energy stored in a spring, which is derived from Hooke's Law. Here’s the step-by-step solution:

### Step 1: Understand Hooke's Law
Hooke's Law states that the force \( F \) exerted by a spring is directly proportional to the extension \( x \) of the spring from its natural length. Mathematically, it is expressed as:
\[ F = kx \]
where \( k \) is the spring constant.

### Step 2: Relate Tension to Extension
In this case, the tension \( T \) in the spring can be considered equivalent to the force \( F \) applied to the spring when it is extended. Therefore, we can write:
\[ T = kx \]

### Step 3: Calculate the Energy Stored
The energy \( U \) stored in the spring when it is extended by \( x \) can be calculated using the formula for elastic potential energy:
\[ U = \frac{1}{2} F x \]
Substituting \( F \) with \( T \) (since \( T = kx \)):
\[ U = \frac{1}{2} T x \]

### Step 4: Substitute for \( x \)
From the relationship \( T = kx \), we can express \( x \) in terms of \( T \):
\[ x = \frac{T}{k} \]
Now substitute this expression for \( x \) back into the energy formula:
\[ U = \frac{1}{2} T \left(\frac{T}{k}\right) \]
This simplifies to:
\[ U = \frac{T^2}{2k} \]

### Conclusion
Thus, the energy stored in the spring when it extends by \( x \) is given by:
\[ U = \frac{T^2}{2k} \]