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(B) The maximum velocity $V_{max}$ of a body performing simple harmonic motion is given by $V_{max} = A \omega$,where $A$ is the amplitude and $\omega

Vedclass · Accepted Answer

$\sqrt{\frac{k_1}{k_2}}$

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(B) The maximum velocity $V_{max}$ of a body performing simple harmonic motion is given by $V_{max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.For a mass $m$ attached to a spring of constant $k$,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$.Given that the masses are equal $(m_A = m_B = m)$ and the amplitudes are equal $(A_A = A_B = A_0)$,the maximum velocities are:$V_A = A_0 \omega_A = A_0 \sqrt{\frac{k_1}{m}}$$V_B = A_0 \omega_B = A_0 \sqrt{\frac{k_2}{m}}$The ratio of the maximum velocity of $A$ to the maximum velocity of $B$ is:$\frac{V_A}{V_B} = \frac{A_0 \sqrt{\frac{k_1}{m}}}{A_0 \sqrt{\frac{k_2}{m}}} = \sqrt{\frac{k_1}{k_2}}$.

Two bodies $A$ and $B$ of equal mass are suspended from two massless springs of spring constant $k_1$ and $k_2$,respectively. If the bodies oscillate vertically such that their amplitudes are equal,the ratio of the maximum velocity of $A$ to the maximum velocity of $B$ is

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