Two equations of two S.H.M. are `y=alpha sin (omegat-alpha) and y=b cos ( omega t -alpha)` . The phase difference between the two is

Options

A. `0^(@)`
B. `alpha^(@)`
C. `90^(@)`
D. `180^(@)`

Answer

To find the phase difference between the two equations of simple harmonic motion (S.H.M.), we will follow these steps:

### Step 1: Identify the equations
We have two equations:
1. \( y_1 = \alpha \sin(\omega t - \alpha) \)
2. \( y_2 = b \cos(\omega t - \alpha) \)

### Step 2: Convert the cosine equation to sine
To compare the two equations, we can convert the cosine function into a sine function. We know that:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_2 \) as:
\[ y_2 = b \cos(\omega t - \alpha) = b \sin\left(\left(\omega t - \alpha\right) + \frac{\pi}{2}\right) \]

### Step 3: Identify the phase angles
Now we can identify the phase angles:
- For \( y_1 \): The phase angle is \( \phi_1 = \omega t - \alpha \)
- For \( y_2 \): The phase angle is \( \phi_2 = \left(\omega t - \alpha + \frac{\pi}{2}\right) \)

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two motions can be calculated as:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the phase angles:
\[ \Delta \phi = \left(\omega t - \alpha + \frac{\pi}{2}\right) - (\omega t - \alpha) \]
Simplifying this gives:
\[ \Delta \phi = \frac{\pi}{2} \]

### Step 5: Convert to degrees
To express the phase difference in degrees, we convert radians to degrees:
\[ \frac{\pi}{2} \text{ radians} = 90^\circ \]

### Final Answer
The phase difference between the two S.H.M. equations is \( 90^\circ \).

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