Question

A particle of mass 200 gm executes S.H.M. The restoring force is provided by a spring of force constant 80 N / m. The time period of oscillations is
A. 0.31 sec
B. 0.15 sec
C. 0.05 sec
D. 0.02 sec

Answer 1

Hint: When a mass is attached to the spring and given an initial displacement, then the spring force arises which makes the mass to oscillate in simple harmonic motion. The time period of the oscillation is proportional to the square root of the ratio of mass and the spring constant.

Formula used:
\[T = 2\pi \sqrt {\dfrac{m}{k}} \], here T is the period of oscillation of the vertical spring-block system.

Complete step by step solution:
The mass of the particle is given as 200 gm
We need to change the mass of the particle to the standard unit.
\[m = \dfrac{{200}}{{1000}}kg\]
\[m = 0.200kg\]

The spring constant of the given spring of the spring-block system is given as 80N/m. So, to find the period of the oscillation of the simple harmonic motion of attached particle when the system is set into motion we use the formula.

Putting the values in the period formula, we get
\[T = 2\pi \sqrt {\dfrac{{0.200}}{{80}}} \sec \]
\[T = 2\pi \sqrt {0.0025} \sec \]
\[T = 0.314s\]
On rounding off the obtained value to the nearest two decimal place, we get the period of oscillation of the spring-block system as 0.31 sec.

Therefore, the correct option is (A).

Note: We should note that the period of oscillation of the spring –block system doesn’t depend on the orientation of the spring. The period will remain the same whether the spring is vertical or on the horizontal surface. But we should be aware of the fact the medium is frictionless.