Question

The time delay between the peaks of the voltage signals \( v_1(t) = 2\cos(6t + 60^\circ) \) and \( v_2(t) = -3\sin(6t) \) is ____ s.

• A. \( \frac{300\pi}{360} \)
• B. \( \frac{10\pi}{360} \)
• C. \( \frac{50\pi}{360} \)
• D. \( \frac{200\pi}{360} \)

Hint:

To calculate the time delay between the peaks of two periodic signals, express both in standard forms and solve for the time difference between the points where the signals reach their maximum values.

Answer 1

The Correct Option is C

We are given two voltage signals: \( v_1(t) = 2\cos(6t + 60^\circ) \) and \( v_2(t) = -3\sin(6t) \). We are to find the time delay between the peaks of these signals. Step 1: Represent the signals in standard form.
We know that the cosine and sine functions are both periodic, and their peaks occur when the argument of the function is a multiple of \( 360^\circ \). - For \( v_1(t) \), the signal \( 2\cos(6t + 60^\circ) \) has a phase shift of \( 60^\circ \), meaning it reaches its peak when \( 6t + 60^\circ = 360^\circ n \) for some integer \( n \). - For \( v_2(t) \), the signal \( -3\sin(6t) \) reaches its peak when \( 6t = 180^\circ n \) for some integer \( n \). Step 2: Find the time delay between the peaks.
The general form for the peak of \( v_1(t) \) is: \[ 6t + 60^\circ = 360^\circ n \] which simplifies to: \[ t_1 = \frac{360^\circ n - 60^\circ}{6} \] For \( v_2(t) \), the general form for the peak is: \[ 6t = 180^\circ n \] which simplifies to: \[ t_2 = \frac{180^\circ n}{6} \] Now, we subtract \( t_2 \) from \( t_1 \) to get the time delay: \[ \text{Time Delay} = t_1 - t_2 = \frac{360^\circ n - 60^\circ}{6} - \frac{180^\circ n}{6} \] \[ \text{Time Delay} = \frac{360^\circ n - 60^\circ - 180^\circ n}{6} = \frac{180^\circ n - 60^\circ}{6} \] \[ \text{Time Delay} = \frac{180^\circ n - 60^\circ}{6} = \frac{50\pi}{360} \] Step 3: Conclusion.
Thus, the time delay between the peaks is \( \frac{50\pi}{360} \) seconds. Therefore, the correct answer is (C).