Question

Consider the motion of a particle along the \(x\)-axis. The position of the particle varies with time \(t\) as \(x(t) = \sin^2(\omega t) \cos^3(\omega t)\), where \(\omega\) is a constant. What is the time period of the motion?

• A. \(\frac{2\pi}{\omega}\)
• B. \(\frac{2\pi}{3\omega}\)
• C. \(\frac{2\pi}{5\omega}\)
• D. \(\frac{2\pi}{15\omega}\)

Hint:

To find the period of a product of trigonometric functions, identify the period of each term.
If one term has period \(\pi/\omega\) and the other has \(2\pi/\omega\), the fundamental period of their product is the LCM, which is \(2\pi/\omega\).

Answer 1

The Correct Option is A

The problem involves determining the time period of a motion described by the position function of a particle moving along the \(x\)-axis. The position as a function of time is given by:

\(x(t) = \sin^2(\omega t) \cos^3(\omega t)\)

1. The first step is to observe any periodic functions involved in the expression. Here, we have trigonometric functions \(\sin(\omega t)\) and \(\cos(\omega t)\), both of which have a fundamental period of \(\frac{2\pi}{\omega}\).
2. The given expression can be rewritten using a trigonometric identity:

\(\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}\) and \(\cos^3(\omega t) = \cos(\omega t) \cdot \cos^2(\omega t) = \cos(\omega t) \cdot \frac{1 + \cos(2\omega t)}{2}\).

1. Since the fundamental period of both \(\sin(\omega t)\) and \(\cos(\omega t)\) is \(\frac{2\pi}{\omega}\), the function \(x(t)\) will repeat its value after a change in \(t\) by an entire period of \(\frac{2\pi}{\omega}\). This is due to the fact that all terms in the product \(x(t)\) include integer multiples of \(\omega t\).
2. Therefore, the time period of \(x(t) = \sin^2(\omega t) \cos^3(\omega t)\) remains \(\frac{2\pi}{\omega}\).

Hence, the correct answer is \(\frac{2\pi}{\omega}\).