(C) When two springs with force constants $K_1$ and $K_2$ are connected in series,the equivalent force constant $K_{eq}$ is given by the formula:$\fra

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$T = 2\pi \sqrt{\frac{m(K_1 + K_2)}{K_1 K_2}}$

(C) When two springs with force constants $K_1$ and $K_2$ are connected in series,the equivalent force constant $K_{eq}$ is given by the formula:$\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$$\frac{1}{K_{eq}} = \frac{K_1 + K_2}{K_1 K_2}$$K_{eq} = \frac{K_1 K_2}{K_1 + K_2}$The time period $T$ of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{K_{eq}}}$.Substituting the value of $K_{eq}$:$T = 2\pi \sqrt{\frac{m}{\frac{K_1 K_2}{K_1 + K_2}}}$$T = 2\pi \sqrt{\frac{m(K_1 + K_2)}{K_1 K_2}}$

Class 11 Physics Oscillations SHM of Spring Mass System

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$A$ mass $m$ is suspended from two coupled springs connected in series. The force constants for the springs are $K_1$ and $K_2$. The time period of the suspended mass will be:

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A
$T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$
B
$T = 2\pi \sqrt{\frac{m(K_1 + K_2)}{K_1 K_2}}$
C
$T = 2\pi \sqrt{\frac{m(K_1 + K_2)}{K_1 K_2}}$
D
$T = 2\pi \sqrt{\frac{m K_1 K_2}{K_1 + K_2}}$

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On a smooth inclined plane,a body of mass $M$ is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has a force constant $K$,the period of oscillation of the body (assuming the springs are massless) is

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$A$ mass $m$ is suspended from two coupled springs connected in series. The force constants for the springs are $K_1$ and $K_2$. The time period of the suspended mass will be:

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Three masses $700 \, g, 500 \, g$,and $400 \, g$ are suspended at the end of a spring as shown and are in equilibrium. When the $700 \, g$ mass is removed,the system oscillates with a period of $3 \, s$. When the $500 \, g$ mass is also removed,it will oscillate with a period of ...... $s$.

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