Simple Harmonic Motion Ques 34
- Two bodies $M$ and $N$ of equal masses are suspended from two separate massless springs of spring constants $k _1$ and $k _2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the one amplitude of vibration of $M$ to that of $N$ is
(a) $k _1 / k _2$
(b) $\sqrt{k _2 / k _1}$
(c) $k _2 / k _1$
(d) $\sqrt{k _1 / k _2}$
(1988, 1M)
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Answer:
Correct Answer: 34.(b)
Solution:
Formula:
- $\left(v _M\right) _{\max }=\left(v _N\right) _{\max } $
$\therefore \quad \omega _M A _M=\omega _N A _N$
or $\quad \frac{A _M}{A _N}=\frac{\omega _N}{\omega _M}=\sqrt{\frac{k _2}{k _1}}$
$(\because \omega=\sqrt{\frac{k}{m}})$
$\therefore$ Correct answer is (b).
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