For a simple harmonic motion in a mass-spring | vedclass

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(B) The angular frequency $\omega$ of a mass-spring system is given by the formula $\omega = \sqrt{\frac{k}{m}}$,where $k$ is the spring constant and

Vedclass · Accepted Answer

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

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(B) The angular frequency $\omega$ of a mass-spring system is given by the formula $\omega = \sqrt{\frac{k}{m}}$,where $k$ is the spring constant and $m$ is the mass of the block.Given that the spring constant $k$ remains the same for both cases,we have:$\omega_1 = \sqrt{\frac{k}{m_1}}$ and $\omega_2 = \sqrt{\frac{k}{m_2}}$Taking the ratio of $\omega_2$ to $\omega_1$:$\frac{\omega_2}{\omega_1} = \frac{\sqrt{k/m_2}}{\sqrt{k/m_1}} = \sqrt{\frac{m_1}{m_2}}$Substituting the given values $m_1 = 1\,kg$ and $m_2 = 2\,kg$:$\frac{\omega_2}{\omega_1} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$

For a simple harmonic motion in a mass-spring system shown,the surface is frictionless. When the mass of the block is $1\,kg$,the angular frequency is $\omega_1$. When the mass of the block is $2\,kg$,the angular frequency is $\omega_2$. The ratio $\omega_2 / \omega_1$ is

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