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(D) 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

$\sqrt{\frac{K_{2}}{K_{1}}}$

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(D) 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 $\omega = \sqrt{\frac{k}{m}}$.Given that the maximum velocities of particles $A$ and $B$ are equal,we have $v_{max, A} = v_{max, B}$.Therefore,$A_{1}\omega_{1} = A_{2}\omega_{2}$.Substituting the expression for angular frequency: $A_{1}\sqrt{\frac{K_{1}}{m}} = A_{2}\sqrt{\frac{K_{2}}{m}}$.Since the masses $m$ are equal,they cancel out: $A_{1}\sqrt{K_{1}} = A_{2}\sqrt{K_{2}}$.Rearranging to find the ratio of amplitudes $\frac{A_{1}}{A_{2}}$,we get $\frac{A_{1}}{A_{2}} = \sqrt{\frac{K_{2}}{K_{1}}}$.Thus,the ratio of the amplitude of $A$ to $B$ is $\sqrt{\frac{K_{2}}{K_{1}}}$.

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 the amplitude of $A$ and $B$ is

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