MCQ
The potential energy of a particle $\left(U_x\right)$ executing $S.H.M$. is given by
- ✓$U_x=\frac{k}{2}(x-a)^2$
- B$U_x=k_1 x+k_2 x^2+k_3 x^3$
- C$U_x=A e^{-b x}$
- D$U_x=a\,constant$
✓
Answer
Correct option: A.
$U_x=\frac{k}{2}(x-a)^2$
a
(a)
(a)
$P.E.$ of body in S.H.M. at an instant, $U =\frac{1}{2} m \omega^2 y ^2=\frac{1}{2} ky ^2$
If the displacement, $y=(a-x)$ then
$U =\frac{1}{2} k ( a - x )^2=\frac{1}{2} k ( x - a )^2$
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