Question

A particle of mass $200g$ executes a simple harmonic motion. The restoring force is provided by a spring of spring constant $80N{m^{ - 1}}$. The time period of its motion is
(A) $0.31s$
(B) $0.41s$
(C) $0.62s$
(D) $0.11s$

Answer 1

Hint: Here will use the general relation between time period, mass, and spring constant for a body executing simple harmonic motion to find the exact value of the time period of the given particle executing simple harmonic motion.

Formula used:
For a body that is executing simple harmonic motion, the time period is given by:
$T = 2\pi \sqrt {\dfrac{m}{k}} $
where,
T is the time period of the particle
m is the mass of the particle executing simple harmonic motion
k is the spring constant of the spring
$\pi = 3.14$ a universal constant

Complete answer:
According to the question, we have given that a particle is executing a simple harmonic motion whose mass is given by $m = 200g = 0.2kg$ and the spring constant of the spring is given by $k = 80N{m^1}$

To calculate the time period of the particle executing simple harmonic motion, we can use the general formula as:
$T = 2\pi \sqrt {\dfrac{m}{k}} $

On substituting the values of all the parameters and solving for T in the units of seconds we get
$
  T = 2(3.14)\sqrt {\dfrac{{0.2}}{{80}}} \\
  T = 6.28 \times 0.05 \\
  T = 0.31s \\
 $
So, the time period of the particle is $0.31s$

Hence, the correct answer is option (A) $0.31s$

Note: A simple harmonic motion is one where the particle moves from its means position to an extreme position and comes back again and again with the definite and fixed time period for example swinging of a pendulum and generally the simple harmonic motions are represented by sine and cosine functions as they are periodic in nature.