A mass m is suspended from the two coupled springs connected in series. The force constant for springs are $K_{1}$ and $K_{2}$ . The time period of the suspended mass will be

T = $2 π \sqrt{(\frac{m}{K_{1} + K_{2}})}$

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T = $2 π \sqrt{(\frac{m}{K_{1} + K_{2}})}$

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T = $2 π \sqrt{(\frac{m (K_{1} + K_{2})}{K_{1} K_{2}})}$

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T = $2 π \sqrt{(\frac{m K_{1} K_{2}}{K_{1} + K_{2}})}$

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Solution

The correct option is C
T = $2 π \sqrt{(\frac{m (K_{1} + K_{2})}{K_{1} K_{2}})}$

In series $k_{e q} = \frac{k_{1} k_{2}}{k_{1} + k_{2}}$ so time period T = $2 π \sqrt{\frac{m (k_{1} + k_{2})}{k_{1} k_{2}}}$

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A mass m is suspended from the two coupled springs connected in series. The force constant for springs are K1 and K2. The time period of the suspended mass will be

The correct option is C T = 2π√(m(K1+K2)K1K2) In series keq=k1k2k1+k2 so time period T = 2π√m(k1+k2)k1k2

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are $K_{1}$ and $K_{2}$ . The time period of the suspended mass will be

The correct option is C
T = $2 π \sqrt{(\frac{m (K_{1} + K_{2})}{K_{1} K_{2}})}$

In series $k_{e q} = \frac{k_{1} k_{2}}{k_{1} + k_{2}}$ so time period T = $2 π \sqrt{\frac{m (k_{1} + k_{2})}{k_{1} k_{2}}}$