Question

Two bodies A and B of equal mass are suspended from two separate massles springs of force constant k₁ and k₂, respectively. The bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitudes of body A to that of body B is ______.

Options

• ${\displaystyle \sqrt{\frac{{\text{k}}_2}{{\text{k}}_1}}}$

• ${\displaystyle \frac{{\text{k}}_2}{{\text{k}}_1}}$

• ${\displaystyle \frac{{\text{k}}_1}{{\text{k}}_2}}$

• ${\displaystyle \sqrt{\frac{{\text{k}}_1}{{\text{k}}_2}}}$

Answer 1

Two bodies A and B of equal mass are suspended from two separate massles springs of force constant k₁ and k₂, respectively. The bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitudes of body A to that of body B is ${\displaystyle \underline{\sqrt{\frac{{\text{k}}_2}{{\text{k}}_1}}}}$.

Explanation:

The maximum velocity of body in SHM,

v = Aω

where, A is amplitude of body and ω is the angular frequency.

It is given that, the maximum velocities of bodies are equal.

${\displaystyle {\text{v}}_{\text{A}}={\text{v}}_{\text{B}}}$

${\displaystyle {\text{A}}_{\text{A}}{\omega }_{\text{A}}={\text{A}}_{\text{B}}{\omega }_{\text{B}}}$

${\displaystyle {\text{A}}_{\text{A}}\sqrt{\frac{{\text{k}}_1}{\text{m}}}={\text{A}}_{\text{B}}\sqrt{\frac{{\text{k}}_2}{\text{m}}} ...(\because \omega =\sqrt{\frac{\text{k}}{\text{m}}})}$

${\displaystyle \frac{{\text{A}}_{\text{A}}}{{\text{A}}_{\text{B}}}=\sqrt{\frac{{\text{k}}_2}{{\text{k}}_1}}}$