Question

A spring with force constant $k$ is initially stretched by $x_1$ If it is further stretched by $x_2$, then the increase in its potential energy is

• A. $\frac{1}{2} k(x_2 - x_1)^2$
• B. $\frac{1}{2} k(x_2 + 2x_1)$
• C. $\frac{1}{2} k x_1^2 - \frac{1}{2} k x_2^2$
• D. $\frac{1}{2} k(x_1 + x_2)^2$
• E. $\frac{1}{2} k(x_1^2 + x_2)^2$

Answer 1

Answer: (B)

According to the question, in the first condition
$U_{1}=\frac{1}{2} k x_{1}^{2}$ ...(i)
In the second condition,
$U_{2}=\frac{1}{2} k\left(x_{1}+x_{2}\right)^{2}$ ...(ii)
Increment in the potential energy
$=U_{2}-U_{1}$
$=\frac{1}{2} k\left(x_{1}+x_{2}\right)^{2}-\frac{1}{2} k x_{1}^{2}$
$=\frac{1}{2} k\left(x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}\right)-\frac{1}{2} k x_{1}^{2}$
$=\frac{1}{2} k\left(x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}-x_{1}^{2}\right)$
$=\frac{1}{2} k\left[2 x_{1} x_{2}+x_{2}^{2}\right]$
$=\frac{1}{2} k x_{2}\left(x_{2}+2 x_{1}\right)$