Question

The displacement of a particle executing simple harmonic motion is \( y = A \sin(2\pi t + \phi) \, \text{m} \), where \( t \) is time in seconds and \( \phi \) is the phase angle. At time \( t = 0 \), the displacement and velocity of the particle are 2 m and 4 ms^-1. The phase angle, \( \phi \) =

• A. 60\(^\circ\)
• B. 30\(^\circ\)
• C. 45\(^\circ\)
• D. 90\(^\circ\)

Hint:

In simple harmonic motion, use the displacement and velocity equations at \( t = 0 \) to solve for the phase angle \( \phi \).

Answer 1

The Correct Option is C

Step 1: The general equation for displacement in simple harmonic motion is: \[ y = A \sin(2 \pi t + \phi) \] where \( A \) is the amplitude, \( t \) is the time, and \( \phi \) is the phase angle. At \( t = 0 \), the displacement \( y = 2 \, \text{m} \). Thus, at \( t = 0 \), we have: \[ y = A \sin(\phi) = 2 \] This gives us the first equation: \[ A \sin(\phi) = 2 \quad \text{(Equation 1)} \] Step 2: The velocity in simple harmonic motion is given by: \[ v = A \cdot 2 \pi \cdot \cos(2 \pi t + \phi) \] At \( t = 0 \), the velocity is \( v = 4 \, \text{ms}^{-1} \), so: \[ v = A \cdot 2 \pi \cdot \cos(\phi) = 4 \] This gives us the second equation: \[ A \cdot 2 \pi \cdot \cos(\phi) = 4 \quad \text{(Equation 2)} \] Step 3: Now we have two equations to solve: 1. \( A \sin(\phi) = 2 \) 2. \( A \cdot 2 \pi \cdot \cos(\phi) = 4 \) Dividing Equation 2 by Equation 1: \[ \frac{A \cdot 2 \pi \cdot \cos(\phi)}{A \sin(\phi)} = \frac{4}{2} \] \[ \frac{2 \pi \cos(\phi)}{\sin(\phi)} = 2 \] \[ \frac{\cos(\phi)}{\sin(\phi)} = \frac{2}{2 \pi} = \frac{1}{\pi} \] Thus, \( \tan(\phi) = \pi \), so \( \phi \approx 45^\circ \). Thus, the phase angle is \( \phi = 45^\circ \).