Two simple harmonic motions are given by `y_(1) = a sin [((pi)/(2))t + phi]` and `y_(2) = b sin [((2pi)/( 3))t + phi]`. The phase difference between these after `1 s` is

Options

A. (a)zero
B. `(b)pi//2`
C. `(c)pi//4`
D. `(d)pi//6`

Explanation

To find the phase difference between the two simple harmonic motions after 1 second, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the equations of motion**:
We have two simple harmonic motions:
\[ y_1 = a \sin\left(\frac{\pi}{2} t + \phi\right) \]
\[ y_2 = b \sin\left(\frac{2\pi}{3} t + \phi\right) \]

2. **Substitute \( t = 1 \) second into both equations**:
For \( y_1 \):
\[ y_1(1) = a \sin\left(\frac{\pi}{2} \cdot 1 + \phi\right) = a \sin\left(\frac{\pi}{2} + \phi\right) \]
For \( y_2 \):
\[ y_2(1) = b \sin\left(\frac{2\pi}{3} \cdot 1 + \phi\right) = b \sin\left(\frac{2\pi}{3} + \phi\right) \]

3. **Define the phase angles**:
Let:
\[ \phi_1 = \frac{\pi}{2} + \phi \]
\[ \phi_2 = \frac{2\pi}{3} + \phi \]

4. **Calculate the phase difference**:
The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the values we found:
\[ \Delta \phi = \left(\frac{2\pi}{3} + \phi\right) - \left(\frac{\pi}{2} + \phi\right) \]
The \( \phi \) terms cancel out:
\[ \Delta \phi = \frac{2\pi}{3} - \frac{\pi}{2} \]

5. **Find a common denominator and simplify**:
The common denominator of 3 and 2 is 6. Rewrite the fractions:
\[ \Delta \phi = \frac{4\pi}{6} - \frac{3\pi}{6} = \frac{1\pi}{6} \]
Thus, we have:
\[ \Delta \phi = \frac{\pi}{6} \]

### Final Answer:
The phase difference between the two simple harmonic motions after 1 second is \( \frac{\pi}{6} \).

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