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NEET PHYSICSEasy

A spring is stretched by 5 cm5 \text{ cm} by a force 10 N10 \text{ N}. The time period of the oscillations when a mass of 2 kg2 \text{ kg} is suspended by it is:

A

3.14 s3.14 \text{ s}

B

0.628 s0.628 \text{ s}

C

0.0628 s0.0628 \text{ s}

D

6.28 s6.28 \text{ s}

Step-by-Step Solution

  1. Calculate Spring Constant (kk): According to Hooke's Law, the magnitude of the restoring force FF is related to the extension xx by F=kxF = kx . Given Force (FF) = 10 N10 \text{ N} Extension (xx) = 5 cm=0.05 m5 \text{ cm} = 0.05 \text{ m}
  • k=Fx=100.05=200 N/mk = \frac{F}{x} = \frac{10}{0.05} = 200 \text{ N/m}
  1. Calculate Time Period (TT): The time period of a spring-mass system is given by the formula T=2πmkT = 2\pi \sqrt{\frac{m}{k}} . Given Mass (mm) = 2 kg2 \text{ kg} T=2π2200=2π1100T = 2\pi \sqrt{\frac{2}{200}} = 2\pi \sqrt{\frac{1}{100}} T=2π(110)=2×3.1410T = 2\pi \left( \frac{1}{10} \right) = \frac{2 \times 3.14}{10} T=6.2810=0.628 sT = \frac{6.28}{10} = 0.628 \text{ s}
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