Question

The displacements of two particles executing simple harmonic motion are represented as
$y_1 = 2\sin(10t + \theta)$ and $y_2 = 3\cos(10t)$. The phase difference between the velocities of these waves is

• A. $\theta + \dfrac{\pi}{2}$
• B. $-\theta$
• C. $\theta - \dfrac{\pi}{2}$
• D. $\theta$

Hint:

Velocity in SHM leads displacement by a phase of $\dfrac{\pi}{2}$.

Answer 1

The Correct Option is C

Step 1: Differentiate displacements to get velocities.
\[ v_1 = \frac{dy_1}{dt} = 20\cos(10t + \theta) \] \[ v_2 = \frac{dy_2}{dt} = -30\sin(10t) \]

Step 2: Express both velocities in sine or cosine form.
\[ \cos(10t + \theta) = \sin\left(10t + \theta + \frac{\pi}{2}\right) \] \[ -\sin(10t) = \sin\left(10t - \frac{\pi}{2}\right) \]

Step 3: Identify phase difference.
Phase of $v_1 = \theta + \frac{\pi}{2}$
Phase of $v_2 = -\frac{\pi}{2}$

Step 4: Calculate phase difference.
\[ (\theta + \frac{\pi}{2}) - \frac{\pi}{2} = \theta - \frac{\pi}{2} \]

Step 5: Conclusion.
The phase difference between the velocities is $\theta - \dfrac{\pi}{2}$.