Question: 10. 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\,\text{s}
(2) 6.28\,\text{s}
(3) 3.14\,\text{s}
(4) 0.628\,\text{s}
Answer: Option (4)
Explanation:
When a force F stretches a spring by an extension x,
the spring constant k is given by Hooke’s law F = kx.
Given force F = 10\,\text{N} and extension x = 5\,\text{cm} = 0.05\,\text{m}.
So, the spring constant is k = \frac{F}{x} = \frac{10}{0.05} = 200\,\text{N m}^{-1}.
The time period of oscillation of a mass-spring system is given by T = 2\pi\sqrt{\frac{m}{k}}.
Here, mass m = 2\,\text{kg} and spring constant k = 200\,\text{N m}^{-1}.
Substituting the values, T = 2\pi\sqrt{\frac{2}{200}} = 2\pi\sqrt{\frac{1}{100}}.
This gives T = 2\pi \times \frac{1}{10} = \frac{2\pi}{10} = 0.628\,\text{s}.
Therefore, the correct answer is Option (4).