Two S.H.Ms are given by `y_(1) = a sin ((pi)/(2) t + (pi)/(2))` and `y_(2) = b sin ((2pi)/(3) t + (pi)/(2))` . The phase difference between these after 1 second is

Options

A. `pi`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `(pi)/(6)`

Answer

To find the phase difference between the two simple harmonic motions (S.H.Ms) given by the equations:

1. \( y_1 = a \sin\left(\frac{\pi}{2} t + \frac{\pi}{2}\right) \)
2. \( y_2 = b \sin\left(\frac{2\pi}{3} t + \frac{\pi}{2}\right) \)

we will follow these steps:

### Step 1: Determine the phase of \( y_1 \) at \( t = 1 \) second

The phase of the first S.H.M, \( \phi_1 \), is given by:

\[ \phi_1 = \frac{\pi}{2} t + \frac{\pi}{2} \]

Substituting \( t = 1 \):

\[ \phi_1 = \frac{\pi}{2} \cdot 1 + \frac{\pi}{2} = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

### Step 2: Determine the phase of \( y_2 \) at \( t = 1 \) second

The phase of the second S.H.M, \( \phi_2 \), is given by:

\[ \phi_2 = \frac{2\pi}{3} t + \frac{\pi}{2} \]

Substituting \( t = 1 \):

\[ \phi_2 = \frac{2\pi}{3} \cdot 1 + \frac{\pi}{2} = \frac{2\pi}{3} + \frac{\pi}{2} \]

To add these fractions, we need a common denominator. The least common multiple of 3 and 2 is 6.

Converting each term:

\[ \frac{2\pi}{3} = \frac{4\pi}{6} \]
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]

Now, adding them:

\[ \phi_2 = \frac{4\pi}{6} + \frac{3\pi}{6} = \frac{7\pi}{6} \]

### Step 3: Calculate the phase difference

The phase difference \( \Delta \phi \) is given by:

\[ \Delta \phi = \phi_2 - \phi_1 \]

Substituting the values we found:

\[ \Delta \phi = \frac{7\pi}{6} - \pi \]

Converting \( \pi \) to sixths:

\[ \pi = \frac{6\pi}{6} \]

Thus,

\[ \Delta \phi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6} \]

### Final Answer

The phase difference between the two S.H.Ms after 1 second is:

\[ \Delta \phi = \frac{\pi}{6} \]

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