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(C) The maximum velocity of a particle in simple harmonic motion is given by $v_{max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angula

Vedclass · Accepted Answer

$\sqrt{k_2/k_1}$

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(C) The maximum velocity of a particle in simple harmonic motion is given by $v_{max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.For a spring-mass system,the angular frequency is $\omega = \sqrt{k/m}$.Given that the masses are equal $(m_1 = m_2 = m)$,the angular frequencies are $\omega_1 = \sqrt{k_1/m}$ and $\omega_2 = \sqrt{k_2/m}$.Since the maximum velocities are equal,we have $A_1\omega_1 = A_2\omega_2$.Therefore,the ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.Substituting the expressions for $\omega_1$ and $\omega_2$,we get $\frac{A_1}{A_2} = \frac{\sqrt{k_2/m}}{\sqrt{k_1/m}} = \sqrt{\frac{k_2}{k_1}}$.

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_1$ and $k_2,$ respectively. If the maximum velocities during oscillations are equal,the ratio of amplitudes of $A$ and $B$ is

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