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Physics-

General

Easy

Question

Two springs are joined and attached to a mass of 16 kg. This system is then suspended vertically from a rigid support. The spring constant of the two springs are $k subscript 1 end subscript a n d k subscript 2 end subscript$ respectively. The period of vertical oscillations of the system will be

$\frac{1}{8 π} \sqrt{k_{1} + k_{2}}$
$8 π \sqrt{\frac{k_{1} + k_{2}}{k_{1} k_{2}}}$
$\frac{π}{2} \sqrt{k_{1} - k_{2}}$
$\frac{π}{2} \sqrt{\frac{k_{1}}{k_{2}}}$

The correct answer is: $8 pi square root of fraction numerator k subscript 1 end subscript plus k subscript 2 end subscript over denominator k subscript 1 end subscript k subscript 2 end subscript end fraction end root$

The two springs are in series. Therefore, the time period is
$T equals 2 pi square root of fraction numerator m over denominator k end fraction end root$
$equals blank 2 pi square root of m open parentheses fraction numerator k subscript 1 end subscript plus k subscript 2 end subscript over denominator k subscript 1 end subscript k subscript 2 end subscript end fraction close parentheses end root$
As $m equals 16$ kg;
$T equals 8 pi square root of fraction numerator k subscript 1 end subscript plus k subscript 2 end subscript over denominator k subscript 1 end subscript k subscript 2 end subscript end fraction end root$

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