Two massless springs with spring constants 2k | vedclass

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(B) The maximum velocity of a simple harmonic oscillator is given by $V_{\max} = \omega A$,where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequenc

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$3: 2$

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(B) The maximum velocity of a simple harmonic oscillator is given by $V_{\max} = \omega A$,where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency and $A$ is the amplitude.Given $V_{\max,1} = V_{\max,2}$,we have $\omega_1 A_1 = \omega_2 A_2$,which implies $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.Substituting $\omega = \sqrt{\frac{k}{m}}$,we get $\frac{A_1}{A_2} = \sqrt{\frac{k_2}{m_2}} 	imes \sqrt{\frac{m_1}{k_1}} = \sqrt{\frac{k_2}{k_1} 	imes \frac{m_1}{m_2}}$.Given $k_1 = 2k$,$k_2 = 9k$,$m_1 = 50\,g$,and $m_2 = 100\,g$.Thus,$\frac{A_1}{A_2} = \sqrt{\frac{9k}{2k} 	imes \frac{50}{100}} = \sqrt{\frac{9}{2} 	imes \frac{1}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.

Two massless springs with spring constants $2k$ and $9k$ carry $50\,g$ and $100\,g$ masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then,the ratio of their respective amplitudes will be

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