Question

The function of time representing a simple harmonic motion with a period of \(\frac{\pi}{\omega}\) is:

• A. \(\sin(\omega t) + \cos(\omega t)\)
• B. \(\sin^2(\omega t)\)
• C. \(3\cos\left(\frac{\pi}{4} - 2\omega t\right)\)
• D. \(\cos(\omega t) + \cos(2\omega t) + \cos(3\omega t)\)

Hint:

In SHM equations, the coefficient of \(t\) is the angular frequency \(\Omega\). Always check if \(T = 2\pi/\Omega\) matches your target period.

Answer 1

The Correct Option is C

Step 1: For a function to represent SHM, it must be in the form \(y = A\sin(\Omega t + \phi)\) or \(y = A\cos(\Omega t + \phi)\).
Step 2: The time period \(T\) is related to the angular frequency \(\Omega\) by \(T = \frac{2\pi}{\Omega}\).
Step 3: Given \(T = \frac{\pi}{\omega}\). Setting these equal: \(\frac{2\pi}{\Omega} = \frac{\pi}{\omega} \implies \Omega = 2\omega\).
Step 4: Examine the options. Option (C) has the form \(3\cos(\phi - \Omega t)\) where \(\Omega = 2\omega\). This perfectly matches the required periodicity and SHM form. Note: \(\sin^2(\omega t)\) is periodic but represents SHM about a shifted mean position with frequency \(2\omega\), however, (C) is the most direct standard representation.