Check the correctness of the equation : `y = a "sin" (omega t + phi)`, where y = displacement , a = amplitude , `omega` = angular frequency and `phi` is an angle .

Answer

To check the correctness of the equation \( y = a \sin(\omega t + \phi) \), we need to analyze the dimensions of each term in the equation.

### Step-by-Step Solution:

1. **Identify the Variables**:
- \( y \): Displacement
- \( a \): Amplitude
- \( \omega \): Angular frequency
- \( t \): Time
- \( \phi \): Phase angle

2. **Determine the Dimensions**:
- The dimension of displacement \( y \) is length, which is represented as:
\[ [y] = L^1 \]
- The amplitude \( a \) is also a measure of length, so:
\[ [a] = L^1 \]
- The angular frequency \( \omega \) is defined as:
\[ \omega = \frac{\theta}{t} \]
where \( \theta \) is dimensionless (angle). Thus, the dimension of \( \omega \) is:
\[ [\omega] = T^{-1} \]
- The time \( t \) has the dimension:
\[ [t] = T^1 \]
- The phase angle \( \phi \) is also dimensionless:
\[ [\phi] = L^0 M^0 T^0 \]

3. **Analyze the Right-Hand Side (RHS)**:
- The RHS of the equation is \( a \sin(\omega t + \phi) \).
- Since \( \sin \) is a trigonometric function, its argument must be dimensionless. Therefore, we need to check the dimensions of \( \omega t + \phi \):
- The dimension of \( \omega t \) is:
\[ [\omega] \cdot [t] = T^{-1} \cdot T^1 = L^0 M^0 T^0 \quad \text{(dimensionless)} \]
- Since \( \phi \) is also dimensionless, \( \omega t + \phi \) is dimensionless.

4. **Combine the Dimensions**:
- Now we can express the RHS:
\[ [RHS] = [a] \cdot [\sin(\omega t + \phi)] = L^1 \cdot L^0 M^0 T^0 = L^1 \]

5. **Compare Dimensions**:
- Now, we compare the dimensions of the LHS and RHS:
- LHS: \( [y] = L^1 \)
- RHS: \( [a \sin(\omega t + \phi)] = L^1 \)

Since both sides of the equation have the same dimensions, we conclude that the equation \( y = a \sin(\omega t + \phi) \) is indeed correct.