Question

When a spring is compressed by $3\,cm$, the potential energy stored in it is $U$. When it is compressed further by $6\,cm$, the increase in potential energy is,
A. $4U$
B. $U$
C. $2U$
D. $3U$

Answer 1

Hint:We need to understand spring motion to answer this question. Using the equation of Elastic Potential Energy (relation between energy of spring, displacement in spring and spring constant) we can achieve energy in spring at any specific amount of compression in spring. Evaluate change in potential energy using the equation of elastic potential energy and compare the values.

Formula used:
$P.E. = \dfrac{1}{2}k{x^2}$
Where, $P.E.$ = Potential Energy stored in system, $k$ = spring constant and $x$ = displacement from equilibrium position.

Complete step by step answer:
Spring system, when shift from its equilibrium position will store some energy in form of potential energy which can be obtained as
$P.E. = \dfrac{1}{2}k{x^2}$
Now according to the question when spring is displaced by 3 cm from its equilibrium position it produces potential energy of U. Let spring constant of spring = $k$. Putting these values in equation will give,
$U = \dfrac{1}{2}k{\left( 3 \right)^2}$
$\Rightarrow U = \dfrac{9}{2}k$ …………….(i)

Now spring is more displaced by 6 cm more from 3 cm, i.e. total displacement from mean position is given as 9 cm. And the change in displacement is 6 cm. Let change in potential energy be $\Delta P$. Putting value in equation,
$\Delta P = \dfrac{1}{2}k{\left( 6 \right)^2}$
$\Rightarrow \Delta P = \dfrac{{36}}{2}k$

We can write change in potential energy as,
$\Delta P = \dfrac{{4 \times 9}}{2}k$ ………….(ii)
Comparing equation (i) with equation (ii),
$\Delta P = 4 \times \dfrac{9}{2}k$
From equation (i), we know that $U = \dfrac{9}{2}k$ , hence
$\Delta P = 4U$
So on further compression of spring stored Potential energy is changed by $4U$.

Hence, the correct answer is option A.

Note:Approaching by calculating potential energy in spring at different compressions and then subtracting to yield result will lead to a wrong result. This is because at that instant you will be calculating change in potential energy between two compression, i.e. 9 cm and 3 cm. But now it is potential energy when spring is compressed from 3 cm to 9 cm. Always remember that the spring system involves only displacement from initial position and not distance of oscillation of spring.