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NEET PHYSICSMedium

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is

1

0.0628 s

2

6.28 s

3

3.14 s

4

0.628 s

Step-by-Step Solution

For a spring, F=kxF = kx. 10=k⋅0.0510 = k \cdot 0.05, so k=100.05=200 N/mk = \frac{10}{0.05} = 200 \text{ N/m}. Time period T=2πmk=2π2200=2π1100=2π⋅110=2⋅3.1410=0.628 sT = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2}{200}} = 2\pi \sqrt{\frac{1}{100}} = 2\pi \cdot \frac{1}{10} = \frac{2 \cdot 3.14}{10} = 0.628 \text{ s}.

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