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Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_1$ and $k_2$ 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:
A
$\frac{k_2}{k_1}$
B
$\frac{k_1}{k_2}$
C
$\sqrt{\frac{k_2}{k_1}}$
D
$\sqrt{\frac{k_1}{k_2}}$
Explanation
$v_{max} = \omega A = \sqrt{k/m} A$. Equating maximum speeds gives $\sqrt{k_1} A_P = \sqrt{k_2} A_Q \implies A_Q / A_P = \sqrt{k_1 / k_2}$.
Detailed Solution
For simple harmonic oscillation of a mass-spring system, the maximum velocity is $v_{max} = \omega A = \sqrt{\frac{k}{m}} A$. Given that both masses are identical ($m_P = m_Q = m$) and have equal maximum velocities: $v_{max,P} = v_{max,Q} \implies \sqrt{\frac{k_1}{m}} A_P = \sqrt{\frac{k_2}{m}} A_Q$. Rearranging for the ratio of amplitudes: $\frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}}$.