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(B) The period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{K}}$,which implies $T \propto \frac{1}{\sqrt{K}}$ or $T^2 \propto \frac{1}

Vedclass · Accepted Answer

$T^2 = t_1^2 + t_2^2$

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(B) The period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{K}}$,which implies $T \propto \frac{1}{\sqrt{K}}$ or $T^2 \propto \frac{1}{K}$.For two springs with spring constants $K_1$ and $K_2$ connected in series,the equivalent spring constant $K_s$ is given by $\frac{1}{K_s} = \frac{1}{K_1} + \frac{1}{K_2}$.Since $T^2 = \frac{4\pi^2 m}{K}$,we have $\frac{1}{K} = \frac{T^2}{4\pi^2 m}$.Substituting this into the series combination formula:$\frac{T^2}{4\pi^2 m} = \frac{t_1^2}{4\pi^2 m} + \frac{t_2^2}{4\pi^2 m}$.Multiplying by $4\pi^2 m$,we get $T^2 = t_1^2 + t_2^2$.

$A$ particle at the end of a spring executes simple harmonic motion with a period $t_1$,while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$,then

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