Question

Two bodies M₁ and M₂ of equal masses are suspended from two separate massless spring of force constant k₁ and k₂ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of M₁ to that of M₂ is

1. k₁/k₂
2. \(\sqrt {\frac{{{{\rm{k}}_1}}}{{{{\rm{k}}_2}}}} \)
3. k₂/k₁
4. \(\sqrt {\frac{{{{\rm{k}}_2}}}{{{{\rm{k}}_1}}}}\)

Answer 1

Correct Answer - Option 4 : \(\sqrt {\frac{{{{\rm{k}}_2}}}{{{{\rm{k}}_1}}}}\)

As this is a case of simple harmonic motion

Maximum velocity is given by: \({v_{max}}\; = \;A\omega\) where A is the amplitude of oscillation.

\(\omega \; = \;\sqrt {\frac{k}{m}} \;\)

Here \({\left( {{V_{max}}} \right)_1}\; = \;{\left( {{V_{max}}} \right)_2} \Rightarrow {A_1}{\omega _1}\; = \;{A_2}{\omega _2}\)

\(\frac{{{A_1}}}{{{A_2}}}\; = \;\frac{{{\omega _2}}}{{{\omega _1}}}\; = \;\sqrt {\frac{{{k_2}}}{{{m_2}}}.\frac{{{m_1}}}{{{k_1}}}} \; = \;\sqrt {\frac{{{k_2}}}{{{k_1}}}} \;\)