Simple Harmonic Motion Ques 12
- A linear harmonic oscillator of force constant $2 \times 10^{6}$ $N / m$ and amplitude $0.01$ $m$ has a total mechanical energy of $160$ $J$. Its
$(1989,2 M)$
(a) maximum potential energy is $100$ $J$
(b) maximum kinetic energy is $100$ $J$
(c) maximum potential energy is $160$ $J$
(d) maximum potential energy is zero
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Answer:
Correct Answer: 12.$(b, c)$
Solution:
Formula:
Total Mechanical Energy (TME):
- $\frac{1}{2} k A^{2}=\frac{1}{2} \times 2 \times 10^{6} \times\left(10^{-2}\right)^{2}=100 J$
This is basically the energy of oscillation of the particle.
$K, U$ and $E$ at mean position $(x=0)$ and extreme position $(x= \pm A)$.
$ \begin{array}{ll} x=0 & \\ K=100 J=\text { Maximum } & K=0 J \\ U=60 J=\text { Minimum } & U=160 J=\text { Maximum } \\ E=160 J=\text { Constant } & E=160 J=\text { Constant } \end{array} $
$\therefore$ Correct options are (b) and (c).
BETA
