Question

The equations for the displacements of two particles in simple harmonic motion are \( y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right) \) and \( y_2 = 0.1\cos(\pi t) \) respectively. The phase difference between the velocities of the two particles at a time \( t = 0 \) is

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{6} \)
• D. \( \frac{\pi}{3} \)

Hint:

For simple harmonic motion, if the displacement is given by \( y = A\sin(\omega t + \phi) \), the velocity is \( v = \frac{dy}{dt} = A\omega\cos(\omega t + \phi) \). The phase of the velocity is \( (\omega t + \phi) \). When calculating the phase difference between two oscillating quantities, ensure both quantities are expressed with the same trigonometric function (e.g., both sine or both cosine) to correctly identify their phases. If one is sine and the other is cosine, use identities like \( \cos\theta = \sin(\theta + \frac{\pi}{2}) \) or \( \sin\theta = \cos(\theta - \frac{\pi}{2}) \) to convert them before comparing phases. The phase difference is the absolute difference between their phases at a specific time.

Answer 1

The Correct Option is C

Step 1: Find the velocity equations for each particle.
For simple harmonic motion, if displacement is \( y = A \sin(\omega t + \phi) \), then velocity is \( v = \frac{dy}{dt} = A\omega \cos(\omega t + \phi) \). For particle 1:
\( y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right) \)
Here, \( A_1 = 0.1 \), \( \omega_1 = 100\pi \), and \( \phi_1 = \frac{\pi}{3} \).
The velocity \( v_1 = \frac{dy_1}{dt} = 0.1 \cdot (100\pi) \cos\left(100\pi t + \frac{\pi}{3}\right) \) \( v_1 = 10\pi \cos\left(100\pi t + \frac{\pi}{3}\right) \). For particle 2:
\( y_2 = 0.1\cos(\pi t) \)
We can rewrite \( \cos(\theta) \) as \( \sin\left(\theta + \frac{\pi}{2}\right) \).
So, \( y_2 = 0.1\sin\left(\pi t + \frac{\pi}{2}\right) \). Here, \( A_2 = 0.1 \), \( \omega_2 = \pi \), and \( \phi_2 = \frac{\pi}{2} \).
The velocity \( v_2 = \frac{dy_2}{dt} = 0.1 \cdot (\pi) \cos\left(\pi t + \frac{\pi}{2}\right) \)
\( v_2 = 0.1\pi \cos\left(\pi t + \frac{\pi}{2}\right) \).
Step 2: Determine the phase of each velocity at \( t = 0 \).
The general form of velocity for SHM is \( v = V_{max} \cos(\omega t + \Phi) \). The phase is \( (\omega t + \Phi) \).
For \( v_1 \), the phase is \( \Phi_1(t) = 100\pi t + \frac{\pi}{3} \).
At \( t=0 \), \( \Phi_1(0) = 100\pi (0) + \frac{\pi}{3} = \frac{\pi}{3} \). For \( v_2 \), the phase is \( \Phi_2(t) = \pi t + \frac{\pi}{2} \).
At \( t=0 \), \( \Phi_2(0) = \pi (0) + \frac{\pi}{2} = \frac{\pi}{2} \).
Step 3: Calculate the phase difference between the velocities at \( t = 0 \).
The phase difference \( \Delta \Phi = |\Phi_1(0) - \Phi_2(0)| \).
\( \Delta \Phi = \left|\frac{\pi}{3} - \frac{\pi}{2}\right| \) To subtract, find a common denominator (6):
\( \Delta \Phi = \left|\frac{2\pi}{6} - \frac{3\pi}{6}\right| \) \( \Delta \Phi = \left|-\frac{\pi}{6}\right| \)
\( \Delta \Phi = \frac{\pi}{6} \).