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 :

Correct answer: (D) √(k_1/k_2). Given m_P=m_Q Also (V_max )_P= (V_max )_Q ∴ A_P ω_P=A_Q ω_Q A_P √frack_1~m_P=A_Q √frack_2~m_Q [∵ ω=√(k/m) ] ∴ fracA_QA_P=√(k_1/k_2)

Q129 / 180 · NEET 2025 · Physics

Two identical point masses P and Q , suspended from two separate massless springs of spring constants $\mathrm{k}_1$ and $\mathrm{k}_2$, respectively, oscillate vertically. If their maximum speeds are the same, the ratio $\left(A_Q / A_P\right)$ of the amplitude $A_Q$ of mass $Q$ to the amplitude $A_P$ of mass $P$ is :

A
$\frac{\mathrm{k}_2}{\mathrm{k}_1}$
B
$\frac{\mathrm{k}_1}{\mathrm{k}_2}$
C
$\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}$
Correct answer
D
$\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}$

Solution

$\begin{aligned} & \text { Given } \mathrm{m}_{\mathrm{P}}=\mathrm{m}_{\mathrm{Q}} \\ & \text { Also }\left(\mathrm{V}_{\max }\right)_{\mathrm{P}}=\left(\mathrm{V}_{\max }\right)_{\mathrm{Q}} \\ & \therefore \quad \mathrm{A}_{\mathrm{P}} \omega_{\mathrm{P}}=\mathrm{A}_{\mathrm{Q}} \omega_{\mathrm{Q}} \\ & \mathrm{A}_{\mathrm{P}} \sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}_{\mathrm{P}}}}=\mathrm{A}_{\mathrm{Q}} \sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}_{\mathrm{Q}}}}\left[\because \omega=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\right] \\ & \therefore \frac{\mathrm{A}_{\mathrm{Q}}}{\mathrm{A}_{\mathrm{P}}}=\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}\end{aligned}$

Hence, the correct answer is option: 'D'.

Solution

Relevant Questions