Question

A $0.2\,kg$ of mass hangs at the end of a spring. When $0.02\,kg$ more mass is added to the end of the spring, it stretches $7\,cm$ more. If the $0.02\,kg$ mass is removed, what will be the period of vibration of the system?

Answer 1

Hint:Since the force exerted by the spring is usually in the opposite direction of the displacement, it is called a restoring force. Pulling down on a spring causes it to extend downward, causing the spring to exert an upward force.

Complete step by step answer:
The spring force applied upward by the spring on the mass is equal to and contrary to the force of gravity. As a result, a displaced spring, like gravity, may be used to define an applied force. That is, if we need to think about an applying power, we can always imagine it coming from the end of an ideal spring connected to some external physical device. Hence,
Force due to weight = Restoring force of the spring
$ \Rightarrow mg = kx$
Where , $m$ is the extra mass added, $k$ is spring constant and $g$ is the gravity.
It is given in the question that;
$m = 0.02Kg \\
\Rightarrow x = 0.07m \\
\Rightarrow g = 9.8m.{\sec ^{ - 2}} \\ $
Substituting these values to the main equation we get,
$0.02 \times 9.8 = k \times 0.07 \\
\Rightarrow k = \dfrac{{0.02 \times 9.8}}{{0.07}} \\
\Rightarrow k = 2.86N.{M^{ - 1}} \\ $
The time of vibration of a mass m on a spring with a constant of spring constant k can be measured as follows:
$T = 2\pi \sqrt {\dfrac{m}{k}} \\
\Rightarrow T = 2\pi \sqrt {\dfrac{{0.2}}{{2.86}}} \\
\Rightarrow T = 2 \times 3.14 \times \sqrt {\dfrac{{0.2}}{{2.86}}} \\
\therefore T = 1.66\sec \\ $
Hence, the period of vibration of the system is $1.66\,s$.

Note:Rubber bands, bungee cords, the keys and buttons on laptops, elevators, appliances, and toys are all examples of springs that we use on a daily basis. Ultrasonic probes, both for surgical purposes and nondestructive research, are examples of vibrations' beneficial applications.