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A 5 kg weight...

Question

A 5 kg weight is suspended from a spring. The spring stretches by 2 cm/kg. If the spring is stretched and released, its time period will be:

0.0628

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0.628

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6.28

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62.8

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Solution

The correct option is B $0.628$ s
Linear SHM : F $=$ - ks $α = - w^{2} x$
Therefore, $w^{2} = \frac{k}{m}$ $T = 2 π \sqrt{\frac{m}{k}}$
The spring constant can be found out by $k = \frac{1 k g}{2 c m} = \frac{10 N}{0.02 m} = 500 N / m$
$T = 2 π \sqrt{\frac{m}{k}} = 2 \times 3.14 \times \sqrt{\frac{5}{500}} = 0.628 s$
Hence, option B is correct.

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A 5 kg weight is suspended from a spring. The spring stretches by 2 cm/kg. If the spring is stretched and released, its time period will be:

The correct option is B 0.628 sLinear SHM : F = - ks α=−w2x Therefore, w2=km T=2π√mk The spring constant can be found out by k=1kg2cm=10N0.02m=500N/m T=2π√mk=2×3.14×√5500=0.628sHence, option B is correct.