First slide

Simple hormonic motion

Question

Two particles A and B of equal masses are suspended from two mass less springs of spring constant k₁ and k₂respectively. If the maximum velocities during oscillation, are equal, the ratio amplitudes of A and B is

Easy

A

$\frac{k_{1}}{k_{2}}$
B

$\sqrt{(\frac{k_{1}}{k_{2}})}$
C

$\frac{k_{2}}{k_{1}}$
D

$\sqrt{(\frac{k_{2}}{k_{1}})}$

Solution

$In S.H.M. V_{max} = Ïa For particle A, v_{1} = {Ï}_{1} a_{1} For particles B v_{2} = {Ï}_{2} a_{2} Given that v_{1} = v_{2} or {Ï}_{1} a_{1} = {Ï}_{2} a_{2} \frac{a_{1}}{a_{2}} = \frac{{Ï}_{2}}{{Ï}_{1}} = \frac{T_{1}}{T_{2}} (âµ Ï = \frac{2 Ï}{T}) The time period of a spring is given by T = 2 Ï \sqrt{(\frac{M}{k})} or T â \sqrt{(\frac{1}{k})} \frac{a_{1}}{a_{2}} = \sqrt{(\frac{k_{2}}{k_{1}})}$

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