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

Step 1: Understanding the Question:
We need to find the fundamental time period of a complex periodic motion defined by the product of two trigonometric functions: \(\sin^2(\omega t)\) and \(\cos^3(\omega t)\).

Step 2: Key Formula or Approach:

• A function \(f(t)\) is periodic with period \(T\) if:
\[ f(t + T) = f(t) \]

• For component functions:
The period of \(\sin^2(\omega t)\) is \(T_1 = \frac{\pi}{\omega}\).
The period of \(\cos^3(\omega t)\) is \(T_2 = \frac{2\pi}{\omega}\).

• The fundamental period of the product function is the Least Common Multiple (LCM) of the periods of its individual components.

Step 3: Detailed Explanation:

• Let the function be \(x(t) = \sin^2(\omega t) \cos^3(\omega t)\).

• Let us test if \(T = \frac{2\pi}{\omega}\) satisfies the periodicity condition:
\[ x\left(t + \frac{2\pi}{\omega}\right) = \sin^2\left(\omega \left(t + \frac{2\pi}{\omega}\right)\right) \cos^3\left(\omega \left(t + \frac{2\pi}{\omega}\right)\right) \] \[ = \sin^2(\omega t + 2\pi) \cos^3(\omega t + 2\pi) \]
• Since \(\sin(\theta + 2\pi) = \sin(\theta)\) and \(\cos(\theta + 2\pi) = \cos(\theta)\).:
\[ x\left(t + \frac{2\pi}{\omega}\right) = \sin^2(\omega t) \cos^3(\omega t) = x(t) \]
• Let us check if there is a smaller period, such as \(T' = \frac{\pi}{\omega}\).:
\[ x\left(t + \frac{\pi}{\omega}\right) = \sin^2(\omega t + \pi) \cos^3(\omega t + \pi) \]
• Since \(\sin(\theta + \pi) = -\sin(\theta)\) and \(\cos(\theta + \pi) = -\cos(\theta)\).:
\[ \sin^2(\omega t + \pi) = (-\sin(\omega t))^2 = \sin^2(\omega t) \] \[ \cos^3(\omega t + \pi) = (-\cos(\omega t))^3 = -\cos^3(\omega t) \]
• This gives:
\[ x\left(t + \frac{\pi}{\omega}\right) = -\sin^2(\omega t) \cos^3(\omega t) = -x(t) \neq x(t) \]
• Hence, the smallest value of \(T\) that satisfies the periodicity condition is \(\frac{2\pi}{\omega}\).

Step 4: Final Answer:
The time period of the motion is \(\frac{2\pi}{\omega}\).