If a spring extends by x on loading, then energy stored by spring is (if T is tension in spring and

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13.Oscillations

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

A

$\frac{{{T^2}}}{{2x}}$

B

$\frac{{{T^2}}}{{2K}}$

C

$\frac{{2K}}{{{T^2}}}$

D

$\frac{{2{T^2}}}{K}$

Solution

(b) $U = \frac{1}{2}K{x^2}$ but $T = Kx$
So energy stored $ = \frac{1}{2}\frac{{{{(Kx)}^2}}}{K} = \frac{1}{2}\frac{{{T^2}}}{K}$

Standard 11

Physics

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