Question

Displacement of a particle at any instant of time t is y = 5 sin(100\(\pi\)t + \(\phi\)). The frequency of oscillation of the particle is:

• A. 100 Hz
• B. 25 Hz
• C. 200 Hz
• D. 50 Hz

Hint:

Always be careful to distinguish between angular frequency \(\omega\) (in rad/s) and frequency \(f\) (in Hz). The term multiplying \(t\) inside the sine/cosine function is always \(\omega\).

Answer 1

The Correct Option is D

Step 1: Identify the standard equation for simple harmonic motion (SHM). The displacement \(y\) in SHM is given by \(y = A \sin(\omega t + \phi)\), where \(A\) is amplitude, \(\omega\) is angular frequency, and \(\phi\) is the phase angle.
Step 2: Extract the angular frequency \(\omega\) from the given equation. The given equation is \(y = 5 \sin(100\pi t + \phi)\). Comparing this to the standard form, we can see that the angular frequency \(\omega = 100\pi\) rad/s.
Step 3: Convert angular frequency (\(\omega\)) to linear frequency (\(f\)). The relationship between them is \(\omega = 2\pi f\). \[ f = \frac{\omega}{2\pi} \]
Step 4: Calculate the frequency. \[ f = \frac{100\pi}{2\pi} = 50 \, \text{Hz} \] The frequency of oscillation is 50 Hz.