Two bodies M and N of equal masses are suspended from two separate mass less springs of 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 of N is

$\frac{k_{1}}{k_{2}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\frac{k_{2}}{k_{1}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$\sqrt{\frac{k_{2}}{k_{1}}}$

$(V_{M})_{m a x} = (V_{N})_{m a x}$

$∴ ω_{M} A_{M} = ω_{N} A_{N}$

Or $\frac{A_{M}}{A_{N}} = \frac{ω_{N}}{ω_{M}} = \sqrt{\frac{k_{2}}{k_{1}}}$ $(ω = \sqrt{\frac{k}{m}})$

Suggest Corrections

Similar questions

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

Q. 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 M to that of N is-

Q. Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k₁ and k₂ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
(a) k₁/k₂

(b)

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

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

Q. Two bodies M and N of equal mass are suspended from two separate massless spring of force 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 of N is?

Join BYJU'S Learning Program

Two bodies M and N of equal masses are suspended from two separate mass less springs of spring constant k1 and k2 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

The correct option is B √k2k1 (VM)max=(VN)max ∴ωMAM=ωNAN Or AMAN=ωNωM=√k2k1 (ω=√km)

Two bodies M and N of equal masses are suspended from two separate mass less springs of 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 of N is

The correct option is B
$\sqrt{\frac{k_{2}}{k_{1}}}$

$(V_{M})_{m a x} = (V_{N})_{m a x}$

$∴ ω_{M} A_{M} = ω_{N} A_{N}$

Or $\frac{A_{M}}{A_{N}} = \frac{ω_{N}}{ω_{M}} = \sqrt{\frac{k_{2}}{k_{1}}}$ $(ω = \sqrt{\frac{k}{m}})$