The phase difference between two `SHM y_(1) = 10 sin (10 pi t + (pi)/(3))` and `y_(2) = 12 sin (8 pi t + (pi)/(4))`to `t = 0.5s` is

Answer

To find the phase difference between the two simple harmonic motions (SHM) given by:

1. \( y_1 = 10 \sin(10\pi t + \frac{\pi}{3}) \)
2. \( y_2 = 12 \sin(8\pi t + \frac{\pi}{4}) \)

at \( t = 0.5 \) seconds, we will follow these steps:

### Step 1: Substitute \( t = 0.5 \) into both equations

For \( y_1 \):
\[ y_1 = 10 \sin(10\pi(0.5) + \frac{\pi}{3}) = 10 \sin(5\pi + \frac{\pi}{3}) \]

For \( y_2 \):
\[ y_2 = 12 \sin(8\pi(0.5) + \frac{\pi}{4}) = 12 \sin(4\pi + \frac{\pi}{4}) \]

### Step 2: Simplify the arguments of the sine functions

For \( y_1 \):
\[ y_1 = 10 \sin(5\pi + \frac{\pi}{3}) = 10 \sin(\frac{\pi}{3}) \quad \text{(since \( \sin(5\pi + x) = \sin(x) \))} \]
\[ = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \]

For \( y_2 \):
\[ y_2 = 12 \sin(4\pi + \frac{\pi}{4}) = 12 \sin(\frac{\pi}{4}) \quad \text{(since \( \sin(4\pi + x) = \sin(x) \))} \]
\[ = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2} \]

### Step 3: Find the phase angles

The phase angle for \( y_1 \) at \( t = 0.5 \) seconds is:
\[ \phi_1 = 5\pi + \frac{\pi}{3} \]

The phase angle for \( y_2 \) at \( t = 0.5 \) seconds is:
\[ \phi_2 = 4\pi + \frac{\pi}{4} \]

### Step 4: Calculate the phase difference

The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_1 - \phi_2 \]
Substituting the values:
\[ \Delta \phi = \left(5\pi + \frac{\pi}{3}\right) - \left(4\pi + \frac{\pi}{4}\right) \]
\[ = (5\pi - 4\pi) + \left(\frac{\pi}{3} - \frac{\pi}{4}\right) \]
\[ = \pi + \left(\frac{4\pi - 3\pi}{12}\right) = \pi + \frac{\pi}{12} \]
\[ = \pi + \frac{\pi}{12} = \frac{12\pi}{12} + \frac{\pi}{12} = \frac{13\pi}{12} \]

### Step 5: Final result

Thus, the phase difference between the two SHMs at \( t = 0.5 \) seconds is:
\[ \Delta \phi = \frac{13\pi}{12} \]