A mass m is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are $k_{1}$ and $k_{2}$ respectively. When set into vertical vibrations, the period will be

2 π \sqrt{(\frac{m}{k_{1} k_{2}})}

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2 π \sqrt{m (\frac{k_{1}}{k_{2}})}

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2 π \sqrt{(\frac{m}{k_{1} - k_{2}})}

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2 π \sqrt{(\frac{m}{k_{1} + k_{2}})}

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Solution

The correct option is A $2 π \sqrt{(\frac{m}{k_{1} + k_{2}})}$
Given spring system has parallel combination,
so The equivalent spring constant will be:
$k_{e q} = k_{1} + k_{2}$

and time period of oscillation of the system will be:
$T = 2 π \sqrt{\frac{m}{(k_{1} + k_{2})}}$

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A mass m is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are k1 and k2 respectively. When set into vertical vibrations, the period will be

The correct option is A 2π√(mk1+k2)Given spring system has parallel combination, so The equivalent spring constant will be: keq=k1+k2 and time period of oscillation of the system will be:T=2π√m(k1+k2)

A mass m is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are $k_{1}$ and $k_{2}$ respectively. When set into vertical vibrations, the period will be

The correct option is A $2 π \sqrt{(\frac{m}{k_{1} + k_{2}})}$
Given spring system has parallel combination,
so The equivalent spring constant will be:
$k_{e q} = k_{1} + k_{2}$

and time period of oscillation of the system will be:
$T = 2 π \sqrt{\frac{m}{(k_{1} + k_{2})}}$