The period of the vertical oscillation of a load of mass 4kg suspended form a spring is $0.4$ sec. When an additional load of mass 5kg is applied, the period of oscillation is

0.9 s

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0.8 s

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0.7 s

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0.6 s

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Solution

The correct option is C 0.6 s
$T = 2 π \sqrt{\frac{m}{R}}$
$T α \sqrt{m}$
$\frac{T_{1}}{\sqrt{m_{1}}} = \frac{T_{2}}{\sqrt{m_{2}}}$
$\frac{0.4}{\sqrt{4}} = \frac{T_{2}}{\sqrt{4 + 5}}$
$T_{2} = \frac{0.4 \times 3}{2}$
$T_{2} = 0.6 s$

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The period of the vertical oscillation of a load of mass 4kg suspended form a spring is 0.4 sec. When an additional load of mass 5kg is applied, the period of oscillation is

The correct option is C 0.6 sT=2π√mRTα√mT1√m1=T2√m20.4√4=T2√4+5T2=0.4×32T2=0.6 s

The period of the vertical oscillation of a load of mass 4kg suspended form a spring is $0.4$ sec. When an additional load of mass 5kg is applied, the period of oscillation is

The correct option is C 0.6 s
$T = 2 π \sqrt{\frac{m}{R}}$
$T α \sqrt{m}$
$\frac{T_{1}}{\sqrt{m_{1}}} = \frac{T_{2}}{\sqrt{m_{2}}}$
$\frac{0.4}{\sqrt{4}} = \frac{T_{2}}{\sqrt{4 + 5}}$
$T_{2} = \frac{0.4 \times 3}{2}$
$T_{2} = 0.6 s$