First slide

Simple hormonic motion

Question

The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is:

Moderate

A

4 U
B

8 U
C

16 U
D

$\frac{U}{4}$

Solution

Let extension produced in a spring be x initially. In stretched condition spring will have potential energy
$U = \frac{1}{2} {kx}^{2}$
Where k is spring constant or force constant
$∴ \frac{U_{1}}{U_{2}} = \frac{{x^{2}}_{1}}{{x^{2}}_{2}}$ -----------------(i)
Given, $U_{1} = U, x_{1} = 2 cm, x_{2} = 8 cm$
Putting these values in Eq.(i), we have
$\frac{U}{U_{2}} = \frac{{(2)}^{2}}{{(8)}^{2}} = \frac{4}{64} = \frac{1}{16}$
$∴ U_{2} = 16 U$

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