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.

Step-by-step Explanation:

Step 1: Calculate the spring constant ($k$)
According to Hooke's Law, the force ($F$) required to stretch a spring by a distance ($x$) is given by the relation:
$F=k·x$
Where:
• $F=10\text{N}$ is the applied force,
• $x=5\text{cm}=0.05\text{m}$ is the displacement (converted to standard SI units of meters).

Rearranging the formula to solve for the spring constant ($k$):
$k=\frac{F}{x}$
Substituting the given values:
$k=\frac{10}{0.05}=200\text{N/m}$

Step 2: Find the time period of oscillation ($T$)
When a mass ($m$) is suspended from the spring, the time period of its simple harmonic oscillations is given by the formula:
$T=2\pi \sqrt{\frac{m}{k}}$
Given:
• Suspended mass, $m=2\text{kg}$
• Spring constant, $k=200\text{N/m}$

Substitute these values into the time period formula:
$T=2\pi \sqrt{\frac{2}{200}}$
Simplify the expression under the square root:
$T=2\pi \sqrt{\frac{1}{100}}$
Taking the square root of $\frac{1}{100}$:
$T=2\pi ·\frac{1}{10}=\frac{\pi }{5}\text{s}$

Using the value of $\pi \approx 3.14$:
$T\approx \frac{3.14}{5}=0.628\text{s}$