Question

The vertical extension in a light spring by a weight of 1 kg suspended from the wire is 9.8 cm. The period of oscillation
A. $$20\pi$$sec
B. $$2\pi$$sec
C. $${\displaystyle \frac{2\pi }{10}}$$sec
D. $$200\pi$$sec

Answer 1

Hint: When the mass is suspended from the spring, then because of force of gravity acting on the mass, the spring elongates to balance the weight of the suspended mass. The elongation is proportional to the weight of the suspended mass.

Formula used:
$$F=kx$$, here F is the spring force, k is the spring constant and x is change in length of the spring.
$$T=2\pi \sqrt{{\displaystyle \frac{m}{k}}}$$, here T is the period of oscillation of the vertical spring-block system.

Complete step by step solution:
Let the spring constant of the given spring is k
It is given that the extension in the spring on suspending the weight of 1 kg is 9.8 cm
$$x=9.8cm$$
$$x=9.8\times {10}^{-2}m$$

The force of gravity acting on the suspended body is equal to the weight of the body,
So the spring force will be equal to the weight of the suspended body.
$$F=mg$$
$$kx=mg$$
$$k={\displaystyle \frac{mg}{x}}$$

Putting the values, we get
$$k={\displaystyle \frac{1\times 9.8}{9.8\times {10}^{-2}}}N/m$$
$$k=100N/m$$
So, the spring constant of the given spring is $$100N/m$$

Using the formula of period of oscillation,
$$T=2\pi \sqrt{{\displaystyle \frac{1}{100}}}\sec$$
$$T={\displaystyle \frac{2\pi }{10}}\sec$$

So, the period of oscillation of the given vertical spring-block system is $${\displaystyle \frac{2\pi }{10}}\sec$$

Therefore, the correct option is (C).

Note: We should be careful while plugin the given data in the required formula. We need to change the unit to the standard unit before plugin into the formula. As the expression for the spring constant in this case contains the weight, so we should be careful about the value of the acceleration due to gravity.