Two identical point masses P and Q,suspended | vedclass

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(D) Given that the masses are identical,$m_P = m_Q = m$.The maximum speed of a particle in simple harmonic motion is given by $V_{	ext{max}} = A \ome

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$\sqrt{k_1 / k_2}$

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(D) Given that the masses are identical,$m_P = m_Q = m$.The maximum speed of a particle in simple harmonic motion is given by $V_{	ext{max}} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.The angular frequency is given by $\omega = \sqrt{k/m}$.Given that $(V_{	ext{max}})_P = (V_{	ext{max}})_Q$,we have $A_P \omega_P = A_Q \omega_Q$.Substituting the expressions for $\omega$,we get $A_P \sqrt{k_1 / m} = A_Q \sqrt{k_2 / m}$.Since $m$ is the same for both,we can cancel $\sqrt{m}$ from both sides:$A_P \sqrt{k_1} = A_Q \sqrt{k_2}$.Rearranging to find the ratio $A_Q / A_P$,we get $A_Q / A_P = \sqrt{k_1} / \sqrt{k_2} = \sqrt{k_1 / k_2}$.

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

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