Question

The time taken by a particle performing S.H.M to pass from point A to B when its velocities are the same is 2 seconds. After another 2 seconds it returns to B. The time period of oscillation is (in seconds):
A. 2
B. 4
C. 6
D. 8

Answer 1

Hint: Time taken by the body to complete one oscillation is known as time period. It is denoted by T. So, as the equilibrium point is in between both A and B so we calculate time period from equilibrium point to a then again at equilibrium point after that from equilibrium point to B point and then again at first place.

Formula used:
Total time period of oscillation $=2\times (T_{AB}+T_{BA})$
Where:
$T_{AB}=$ time taken to go from a to b
$T_{BA}=$ next time taken to return at b

Complete step by step answer:
As per the given problem, point a and b points are such that they are located at same distances from the equilibrium point. So that the total time period will be multiplied by 2 and here the velocity is same so that not only the directions but also the magnitude will be the same.

Total time period of oscillation $=2\times (T_{AB}+T_{BA})$
 $=2\times (2+2)$
$=8$ second
Hence, the correct answer is option D.

Additional Information
Every oscillatory motion is periodic motion but every periodic motion is not oscillatory motion. Differentiate periodic motion by example like- a swing in motion or a bouncing ball or water wave. In an oscillating system, the force always acts during a direction opposite to the displacement of the particle from the equilibrium point. This force is often constant, or it can vary with time or position, and is named a restoring force. As long because the force obeys the above principle, the resulting motion is oscillatory.

Note:
Oscillatory motion is defined as the to and fro motion of the body about its fixed position. Oscillatory motion is a type of periodic motion but periodic motion is defined as the motion that repeats itself after fixed intervals of time. This fixed interval of time is known as the time period of the periodic motion.