Question

What is the period of oscillation of a mass of $40kg$ on a spring with constant $k=10N{{m}^{-1}}$ ?

Answer 1

Hint: The period of oscillation is the time period after which a periodic motion repeats itself. A spring follows simple harmonic motion which is a special type of periodic motion. The time period of oscillation is related to the mass of the body and the spring constant. Substituting given values in the relation, time period can be calculated.
Formulas used:
$T=2\pi \sqrt{\dfrac{m}{k}}$

Complete answer:
A spring undergoes simple harmonic motion which is a special type of periodic motion wherein, the force is directly proportional to the negative of displacement. Therefore,
$F\propto -x$
Here, $F$ is the force
$x$ is the displacement
 On removing the sign of proportionality, we get,
 $F=-kx$
Here, $k$ is the spring constant which is also the constant of proportionality.
 Given that, a body attached to a spring has mass $40kg$, the spring constant of the given spring is $k=10N{{m}^{-1}}$.
The time period of oscillation is the time taken by a body to complete one oscillation. Its SI unit is seconds (s). It is given by-
$T=2\pi \sqrt{\dfrac{m}{k}}$
Here, $T$ is the time period of oscillation
$m$ is the mass of the body
$k$ is the spring constant
Substituting given values in the above equation, we get,
 $\begin{align}
  & T=2\pi \sqrt{\dfrac{40}{10}} \\
 & \Rightarrow T=4\pi \\
 & \therefore T=2\times 3.14=6.28s \\
\end{align}$
Therefore, the time period of the body attached to the spring is $6.28s$.

Note:
Periodic motion is the motion which repeats itself after fixed intervals of time. These fixed intervals are called time periods of motion. In one oscillation, the body moves from mean position to one extreme position, then to the other extreme position and back to the mean position.