Two identical point masses P and Q, suspended from two separate massless springs of spring constants k1 and k2, respectively, oscillate vertically. If their maximum speeds are the same, the ratio (AQ/AP) of the amplitude AQ of mass Q to the amplitude A_p of mass P is

For a mass‑spring system performing simple harmonic motion, the angular frequency is

$\omega =\sqrt{\frac{k}{m}}$

The instantaneous speed reaches its maximum when the displacement equals the amplitude A, so the maximum speed is

$\mathrm{v₍ₘₐₓ₎}=A\omega$

Let the two identical point masses have the same mass m. Their spring constants are k₁ for mass P and k₂ for mass Q. Their angular frequencies are therefore

$\mathrm{\omega ₍P₎}=\sqrt{\frac{k}{}}$₁m

$\mathrm{\omega ₍Q₎}=\sqrt{\frac{k}{}}$₂m

The problem states that the maximum speeds of the two masses are equal, i.e.

$A$_Pω_P=A_Qω_Q

Solving for the ratio of the amplitudes gives

$\frac{A}{}$_QA_P=ω_Pω_Q

Substituting the expressions for the angular frequencies and cancelling the common mass m, we obtain

$\frac{A}{}$_QA_P=k₁k₂

Thus the required amplitude ratio is

$\sqrt{\frac{k}{}}$₁k₂

This matches the provided correct option.