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 :

The correct answer is 0.628 s.

To find the time period of oscillation when a mass is suspended by the spring, we first need to determine the spring constant ($k$).

According to Hooke's Law, the relationship between the restoring force ($F$) and the elongation ($x$) of the spring is given by:

$F=kx$

Rearranging the formula to solve for the spring constant ($k$):

$k=\frac{F}{x}$

Given:
Force, $F=10\text{N}$
Elongation, $x=5\text{cm}=0.05\text{m}$

Substituting these values to calculate $k$:

$k=\frac{10}{0.05}=200\text{N/m}$

The time period of oscillation ($T$) for a suspended mass ($m$) on a spring is given by the formula:

$T=2\pi \sqrt{\frac{m}{k}}$

Substitute the mass $m=2\text{kg}$ and the spring constant $k=200\text{N/m}$ into the equation:

$T=2\pi \sqrt{\frac{2}{200}}$

$T=2\pi \sqrt{\frac{1}{100}}$

$T=2\pi ·\frac{1}{10}$

$T=\frac{2·3.14}{10}$

$T=\frac{6.28}{10}=0.628\text{s}$