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:
$\sqrt{\frac{k_1}{k_2}}$

Step-by-step derivation:

For a body executing vertical simple harmonic motion (SHM), the maximum velocity (or maximum speed) $v_{\text{max}}$ is given by the relation:
$v_{\text{max}}=A\omega$
where A is the amplitude of the oscillation and $\omega$ is the angular frequency.

For a mass m suspended from a spring of spring constant k, the angular frequency $\omega$ is given by:
$\omega =\sqrt{\frac{k}{m}}$

Substituting this expression for $\omega$ into the formula for maximum speed, we get:
$v_{\text{max}}=A\sqrt{\frac{k}{m}}$

Let both point masses P and Q have the same mass m (since they are identical).
For mass P, the spring constant is $k_1$ and the amplitude is $A_{\text{P}}$. Its maximum speed is:
$v_{\text{max, P}}=A_{\text{P}}\sqrt{\frac{k_1}{m}}$

For mass Q, the spring constant is $k_2$ and the amplitude is $A_{\text{Q}}$. Its maximum speed is:
$v_{\text{max, Q}}=A_{\text{Q}}\sqrt{\frac{k_2}{m}}$

Since it is given that their maximum speeds are equal ($v_{\text{max, P}}=v_{\text{max, Q}}$), we can equate the two expressions:
$A_{\text{P}}\sqrt{\frac{k_1}{m}}=A_{\text{Q}}\sqrt{\frac{k_2}{m}}$

Dividing both sides by the common factor $\sqrt{\frac{1}{m}}$, we obtain:
$A_{\text{P}}\sqrt{k_1}$=AQk2

Rearranging the equation to find the ratio of the amplitude of mass Q to the amplitude of mass P ($\frac{A_{\text{Q}}}{A_{\text{P}}}$):
$\frac{A_{\text{Q}}}{A_{\text{P}}}=\frac{\sqrt{k_1}}{\sqrt{k_2}}$

Thus, the ratio is:
$\frac{A_{\text{Q}}}{A_{\text{P}}}=\sqrt{\frac{k_1}{k_2}}$