The frequency of oscillations of a mass m con | vedclass

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(D) The frequency of a mass-spring system is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.Given,the initial frequency $f_1 = 4 \ Hz$,so $4 = \frac

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$2 \sqrt{2} \ Hz$

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(D) The frequency of a mass-spring system is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.Given,the initial frequency $f_1 = 4 \ Hz$,so $4 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.When two identical springs of constant $k$ are connected in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$,which implies $k_{eq} = \frac{k}{2}$.The new frequency $f_2$ is given by $f_2 = \frac{1}{2\pi} \sqrt{\frac{k_{eq}}{m}} = \frac{1}{2\pi} \sqrt{\frac{k/2}{m}} = \frac{1}{\sqrt{2}} \left( \frac{1}{2\pi} \sqrt{\frac{k}{m}} ight)$.Substituting the value of the initial frequency,$f_2 = \frac{1}{\sqrt{2}} 	imes 4 = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ Hz$.

The frequency of oscillations of a mass $m$ connected horizontally by a spring of spring constant $k$ is $4 \ Hz$. When the spring is replaced by two identical springs connected in series as shown in the figure,the effective frequency is:

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