Question

Two bodies M and N of equal masses are suspended from two separate mass less springs of force constants $k_{1}$and $k_{2}$respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of M to that of N is

• A. $k_{1}/k_{2}$
• B. $\sqrt{k_{1}/k_{2}}$
• C. $k_{2}/k_{1}$
• D. $\sqrt{k_{2}/k_{1}}$

Answer 1

Answer: (D)

$\sqrt{k_{2}/k_{1}}$

Answer 2

Given that maximum velocities are equal $a_{1}\omega_{1} = a_{2}\omega_{2}$

⇒ $a_{1}\sqrt{\frac{k_{1}}{m}} = a_{2}\sqrt{\frac{k_{2}}{m}}$⇒ $\frac{a_{1}}{a_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$.