Question

When a mass m is attached to a spring, it normally extends by 0.2 m. The mass m is given a slight addition extension and released, then its time period will be
A. \[\dfrac{1}{7}\,\sec \]
B. 1 sec
C. \[\dfrac{{2\pi }}{7}\,\sec \]
D. \[\dfrac{2}{{3\pi }}\,\sec \]

Answer 1

Hint:Time period is defined as the time taken by the spring oscillator to complete one oscillation and it is directly proportional to square root of mass of the system and inversely proportional to square root of spring constant.

Formula used :
Spring force \[F = mg = kx\,\,\]and Time period, \[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Here, K = Spring constant, x = Elongation in spring, m = mass of system and g = acceleration due to gravity.

Complete step by step solution:
Given here is a mass (m) attached to a spring which normally extends to 0.2 m, we have to find the time period of the spring mass oscillator system. Spring force F is given by,
\[F = mg = kx\,\,\]
Using spring force equation constant k can be written as,
\[k = \dfrac{{mg}}{x}\,\,.....(1)\]
As we know that time period of spring mass oscillator system having constant k and mass m is given by,
\[T = 2\pi \sqrt {\dfrac{m}{k}} \,\,\,........(2)\]

Substituting the value of constant k from equation (1) in equation (2) we get,
\[T = 2\pi \sqrt {\dfrac{{mx}}{{mg}}} \\
\Rightarrow T = 2\pi \sqrt {\dfrac{x}{g}} \]
Substituting x = 0.2 m and \[g = 10\,m/{S^2}\] in above equation we get,
\[T = 2\pi \sqrt {\dfrac{{0.2}}{{10}}} \\
\Rightarrow T = 2\pi \sqrt {\dfrac{2}{{100}}} \]
Further solving the above equation we get,
\[T = 2\pi \times \dfrac{{1.4}}{{10}} \\
\therefore T = \dfrac{{2\pi }}{7}\]
Hence, the time period of the given spring mass system is \[\dfrac{{2\pi }}{7}\].

Therefore, option C is the correct answer.

Note: Here, mass is cancelled out because there is no change in the mass of the system when spring is given slight additional extension and released.