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 (A_Q/A_P) of the amplitude A_Q of mass Q to the amplitude A_P of mass P is:

The correct answer is $\sqrt{\frac{k_1}{k_2}}$.

Let's understand the step-by-step derivation to reach this solution.

A point mass suspended from a spring undergoes Simple Harmonic Motion (SHM) when displaced. The maximum speed $v_{\text{max}}$ of an object executing simple harmonic motion with amplitude $A$ and angular frequency $\omega$ is given by:
$v_{\text{max}}=A\omega$

The angular frequency $\omega$ of a mass $m$ oscillating on a spring with spring constant $k$ is given by:
$\omega =\sqrt{\frac{k}{m}}$

Thus, we can express the maximum speed of the oscillating mass as:
$v_{\text{max}}=A\sqrt{\frac{k}{m}}$

Let the two identical point masses be $m_P=m_Q=m$.
For mass $P$ suspended from the spring of spring constant $k_1$ with amplitude $A_P$, its maximum speed is:
$v_{P,\text{max}}=A_P\sqrt{\frac{k_1}{m}}$

For mass $Q$ suspended from the spring of spring constant $k_2$ with amplitude $A_Q$, its maximum speed is:
$v_{Q,\text{max}}=A_Q\sqrt{\frac{k_2}{m}}$

According to the question, their maximum speeds are the same ($v_{P,\text{max}}=v_{Q,\text{max}}$):
$A_P\sqrt{\frac{k_1}{m}}=A_Q\sqrt{\frac{k_2}{m}}$

Dividing both sides by $\sqrt{\frac{1}{m}}$ gives:
$A_P\sqrt{k_1}=A_Q\sqrt{k_2}$

Now, we solve for the ratio of the amplitude of mass $Q$ to the amplitude of mass $P$ ($\frac{A_Q}{A_P}$):
$\frac{A_Q}{A_P}=\frac{\sqrt{k_1}}{\sqrt{k_2}}=\sqrt{\frac{k_1}{k_2}}$