Simple Harmonic Motion 4 Question 11

11. 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)

Objective Question II (One or more correct option)

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Answer:

Correct Answer: 11. (b)

Solution:

$$ \left(v _M\right) _{\max }=\left(v _N\right) _{\max } $$

$\therefore \quad \omega _M A _M=\omega _N A _N$

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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