Question

The equation of wave is given by $y = 10 \sin \left(\frac{2\pi t}{30} + \alpha\right)$. If the displacement is 5 cm at t = 0, then the total phase at t = 7.5 s will be $\left[ \sin 30^\circ = \cos 60^\circ = \frac{1}{2}, \cos 30^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2} \right]$

• A. $\frac{\pi}{3}$ rad
• B. $\frac{\pi}{2}$ rad
• C. $\frac{2\pi}{5}$ rad
• D. $\frac{2\pi}{3}$ rad

Hint:

Notice that a time of $t = 7.5\ \text{s}$ is exactly one-quarter of the wave's total time period ($T = 30\ \text{s}$). Moving a quarter-period forward adds exactly $\frac{\pi}{2}$ rad ($90^\circ$) to the initial phase. Adding $90^\circ$ to our starting angle of $30^\circ$ ($\frac{\pi}{6}$) gives $120^\circ$, which is exactly $\frac{2\pi}{3}$ rad!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Given a wave function equation $y(t)$ with an initial condition entry at $t = 0$, we need to compute the absolute total phase angle argument $\phi = \left(\frac{2\pi t}{30} + \alpha\right)$ at time marker $t = 7.5\ \text{s}$.

Step 2: Detailed Explanation:
First, let's solve for the initial phase angle $\alpha$ using the boundary state condition ($y = 5\ \text{cm}$ at $t = 0$): $$ 5 = 10 \sin \left(\frac{2\pi(0)}{30} + \alpha\right) \implies 5 = 10 \sin\alpha $$ $$ \sin\alpha = \frac{5}{10} = \frac{1}{2} \implies \alpha = \frac{\pi}{6}\ \text{rad} $$ Now, we calculate the total phase angle argument $\phi$ at our target time $t = 7.5\ \text{s}$: $$ \phi = \frac{2\pi(7.5)}{30} + \alpha $$ Substitute the exact values into the phase equation: $$ \phi = \frac{15\pi}{30} + \frac{\pi}{6} = \frac{\pi}{2} + \frac{\pi}{6} $$ Taking a common denominator of 6 to complete the angle addition: $$ \phi = \frac{3\pi + \pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}\ \text{rad} $$

Step 3: Final Answer:
The total phase at $t = 7.5\ \text{s}$ is $\frac{2\pi}{3}$ rad, which corresponds to option (D).