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

\frac{K_{2}}{K_{1}}

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\sqrt{\frac{K_{2}}{K_{1}}}

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\sqrt{K_{1} K_{2}}

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\sqrt{\frac{K_{1}}{K_{2}}}

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Solution

The correct option is B $\sqrt{\frac{K_{2}}{K_{1}}}$
In case of SHM, maximum velocity = $a ω$
$∴ v_{M} = a_{M} ω_{M} = \frac{a_{M} \times 2 π}{T_{M}}$
$v_{N} = a_{N} ω_{N} = a_{N} \times \frac{2 π}{T_{N}}$
But $v_{M} = v_{N} (g i v e n)$
$∴ \frac{a_{M} \times 2 π}{T_{M}} = \frac{a_{N} \times 2 π}{T_{N}} o r \frac{a_{M}}{a_{N}} = \frac{T_{M}}{T_{N}} . . . (i)$
As $\frac{T_{M}}{T_{N}} = 2 π \sqrt{\frac{m}{k_{1}}} \times \frac{1}{2 π} \sqrt{\frac{k_{2}}{m}}$
or $\frac{T_{M}}{T_{N}} = \sqrt{\frac{k_{2}}{k_{1}}} . . . (i i)$
From (i) and (ii), $\frac{a_{M}}{a_{N}} = \sqrt{\frac{k_{2}}{k_{1}}}$

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

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