Simple Harmonic Motion Ques 13
- An object of mass $0.2 kg$ executes simple harmonic oscillations along the $X$-axis with a frequency of $(25 / \pi) Hz$. At the position $x=0.04$, the object has kinetic energy of $0.5 J$ and potential energy $0.4 J$. The amplitude of oscillations is …….m.
(1994, 2M)
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Answer:
Correct Answer: 13.0.06
Solution:
Formula:
Total Mechanical Energy (TME):
- Since, $\nu=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
$\Rightarrow \frac{25}{\pi}=\frac{1}{2 \pi} \sqrt{\frac{k}{0.2}}$
or $k=50 \times 50 \times 0.2=500 N / m$
If $A$ is the amplitude of oscillation,
Total energy $=KE+PE$
$$ \begin{aligned} \frac{1}{2} k A^{2} & =0.5+0.4 \\ A & =\sqrt{\frac{2 \times 0.9}{500}} \\ & =0.06 m \end{aligned} $$
BETA
