Two bodies M and N of equal masses are suspended from two separate massless 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 M to that of N is

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Solution

The correct option is D

Maximum velocity = a $ω = a \sqrt{\frac{k}{m}}$

Given that $a_{1} \sqrt{\frac{K_{1}}{m}} = a_{2} \sqrt{\frac{K_{2}}{m}} \Rightarrow \frac{a_{1}}{a_{2}} = \sqrt{\frac{K_{2}}{K_{1}}}$

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Two bodies M and N of equal masses are suspended from two separate massless springs of force constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude M to that of N is

The correct option is D Maximum velocity = a ω=a√km Given that a1√K1m=a2√K2m⇒a1a2=√K2K1

Two bodies M and N of equal masses are suspended from two separate massless 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 M to that of N is

The correct option is D

Maximum velocity = a $ω = a \sqrt{\frac{k}{m}}$

Given that $a_{1} \sqrt{\frac{K_{1}}{m}} = a_{2} \sqrt{\frac{K_{2}}{m}} \Rightarrow \frac{a_{1}}{a_{2}} = \sqrt{\frac{K_{2}}{K_{1}}}$