If spring extends by x on loading , then enrgy stored by the spring is `……….` ( If T is the tension of the spring and K is the spring constant )

Options

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

Answer

B

Explanation

For spring T = kx
` :. X = T/k " " ….(1)`
Now energy stored in spring `U = 1/2 kx^(2)`
`=1/2 k(T/k)^(2)`
` = (T^(2))/(2k)`