Question

The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along

• A. circle
• B. figure of eight
• C. straight line
• D. ellipse

Answer 1

The Correct Option is C

To solve the problem, we first need to understand the combined effect of two simple harmonic motions (SHM) that occur at right angles to each other with a phase difference of $\pi$.

In simple harmonic motion, the displacement of a particle in the horizontal direction can be represented as:

x = A \cos(\omega t)

And the displacement in the vertical direction can be represented as:

y = A \cos(\omega t + \pi) = -A \cos(\omega t)

Since \cos(\omega t + \pi) = -\cos(\omega t) due to the trigonometric identity for cosine, we substitute to find the equation that describes the motion:

Thus, the equations for the horizontal and vertical motion are:

x = A \cos(\omega t)

y = -A \cos(\omega t)

Combining these equations gives us the relationship:

\frac{x}{A} + \frac{y}{A} = \cos(\omega t) - \cos(\omega t) = 0

Therefore, simplifying the above relationship, we get:

x + y = 0

This equation represents a straight line that passes through the origin and has a slope of -1. When plotted in a coordinate system, it is evident that the resultant motion is a straight line making a 45-degree angle with both the positive x-axis and negative y-axis.

Thus, the composition of the two SHMs results in the displacement of the particle along a straight line.

Therefore, the correct answer is: straight line.