The frequency of oscillation of a mass m susp | vedclass

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(C) The frequency of oscillation $v$ of a mass $m$ suspended by a spring with force constant $k$ is given by the formula:$v = \frac{1}{2 \pi} \sqrt{\f

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$2v$

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(C) The frequency of oscillation $v$ of a mass $m$ suspended by a spring with force constant $k$ is given by the formula:$v = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$From this relation,we can see that the frequency is inversely proportional to the square root of the mass:$v \propto \frac{1}{\sqrt{m}}$Let the initial mass be $m_1 = m$ and the initial frequency be $v_1 = v$. The new mass is $m_2 = \frac{m}{4}$.Using the proportionality $v \propto \frac{1}{\sqrt{m}}$,we can write:$\frac{v_2}{v_1} = \sqrt{\frac{m_1}{m_2}}$Substituting the values:$\frac{v_2}{v} = \sqrt{\frac{m}{m/4}} = \sqrt{4} = 2$Therefore,the new frequency is $v_2 = 2v$.

The frequency of oscillation of a mass $m$ suspended by a spring is $v$. If the mass is reduced to one-fourth,what will be the new frequency of oscillation?

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