Displacement between the maximum potential energy (P.E.) position and the maximum energy (K.E.) position for…
Displacement between the maximum potential energy (P.E.) position and the maximum energy (K.E.) position for a particle executing simple harmonic motion is:
$\pm \frac{a}{2}$
$\pm \mathrm{a}$
$\pm 2 \mathrm{a}$
- 1
Solution
The K.E. of pendulum
$\begin{aligned}
& =\mathrm{K} . \mathrm{E} . \\
& =\frac{1}{2} K\left(a^2-y^2\right)
\end{aligned}$
and P.E. of pendulum $=\frac{1}{2} K Y^2$
when $y=0$ (mean position) K.E. is maximum and PE is 0 .
When $y=0$, K.E. is 0 and P.E. is maximum.
$\therefore$ The displacement between position of maximum potential energy and maximum kinetic energy is $\pm a$.
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