SATHEE: Simple Harmonic Motion 4 Question 11

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)

Show Answer

Answer:

Correct Answer: 11. (b)

Solution:

${(v_{M})}_{max} = {(v_{N})}_{max}$

$∴ ω_{M} A_{M} = ω_{N} A_{N}$

or $\frac{A_{M}}{A_{N}} = \frac{ω_{N}}{ω_{M}} = \sqrt{\frac{k_{2}}{k_{1}}}$

$∵ ω = \sqrt{\frac{k}{m}}$

$∴$ Correct answer is (b).

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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 k1 and k2 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

Objective Question II (One or more correct option)

Answer:

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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