How to Approach Problems in Simple Harmonic Motion

We already know what a simple harmonic motion is. Now we will see different techniques to approach a problem in SHM. For example, it can either be visualized as a motion of a particle in a uniform circular motion and taking their projections, or it can be dealt as trigonometric equations where we can find the solution by solving those equation. In this article, we will try to visualize this motion as a uniform circular motion.

Suppose there are two particles A and B executing simple harmonic equation given below:

\(X_A\) = A sin (?t)

\(X_B\) = A sin (?t + \( frac{pi}{3}\) )

We have to calculate the time after which these two particles will meet for the first time.

According to the given equation, the system can be represented as below:

simple harmonic motion

If we analyze the projection of A and B we can say that both will meet somewhere when B is in the fourth quadrant and A is the first quadrant. But since both are rotating with same angular speed so the angle between them will remain the same. So suppose they meet after A has moved an angle ?. It is shown in the figure below:

simple harmonic motion

As A and B have same projection on x-axis so, ?AOX = ?BOX   …. (1)

And we know, ?AOB = \( frac{pi}{3}\)   …..(2)

Using (1) and (2) we have,

?AOX = \( frac{pi}{6}\)

From this we conclude that the angle moved by A is \( frac{pi}{3}\) also its angular speed is ?. So time taken can be calculated as:

t = \( frac{pi}{3?}\) s

Similarly we can say the period of oscillation is \( frac{2 pi}{omega}\)  s.

A similar question could be: after what time they will have same velocity for the first time? For this if we consider the above situation the magnitude of velocity will be equal but their direction is different so velocity cannot be equal. So for this case we analyze we can say that B will be somewhere in the third quadrant and A in fourth quadrant. Also their displacement from the mean position will be same in magnitude.

simple harmonic motion

As the displacement should be equal in magnitude so ?AOY = ?BOY   …..(3)

And we know, ?AOB = \( frac{pi}{3}\)   …..(4)

Using (3) and (4) we get,

?AOY = \( frac{pi}{6}\)

So total angle covered by A = ?- \( frac{pi}{6}\)

= \( frac{7 pi}{6}\)

Time taken = \( frac{5 pi}{6 omega}\) S

We can deal with several other examples of simple harmonic motion. If you have any other doubts contact our mentors at

Practise This Question

Identify the compound which is a Bronsted Lowry base but not an Arrhenius base?