Check the correctness of the equation `y=a"sin"(2pi)/(lamda)(vt-x)` using dimensional analysis where y is the displacement of a particle at a distance x from the origin at time `t, lamda`- wavelength v-wave velocity, a- amplitude.

Explanation

To check the correctness of the equation \( y = A \sin\left(\frac{2\pi}{\lambda}(vt - x)\right) \) using dimensional analysis, we will follow these steps:

### Step 1: Identify the dimensions of each term in the equation

1. **Displacement \( y \)**: The dimension of displacement is length, denoted as \([L]\).
2. **Amplitude \( A \)**: The amplitude is also a measure of length, so its dimension is \([L]\).
3. **Wavelength \( \lambda \)**: Wavelength is the distance between two successive crests of a wave, hence its dimension is also \([L]\).
4. **Wave velocity \( v \)**: The dimension of wave velocity is length per time, denoted as \([L T^{-1}]\).
5. **Distance \( x \)**: The dimension of distance is length, denoted as \([L]\).
6. **Time \( t \)**: The dimension of time is denoted as \([T]\).

### Step 2: Analyze the argument of the sine function

The argument of the sine function is given by:
\[ \frac{2\pi}{\lambda}(vt - x) \]

We need to ensure that this entire expression is dimensionless.

1. **Evaluate \( vt \)**:
- The dimension of \( vt \) is:
\[ [L T^{-1}] \cdot [T] = [L] \]
So, \( vt \) has the dimension of length \([L]\).

2. **Evaluate \( vt - x \)**:
- Since both \( vt \) and \( x \) have the dimension of length \([L]\), their difference \( vt - x \) also has the dimension of length \([L]\).

3. **Combine with \( \lambda \)**:
- Now, we have:
\[ \frac{vt - x}{\lambda} \]
The dimension of this term is:
\[ \frac{[L]}{[L]} = [1] \]
This shows that \( vt - x \) divided by \( \lambda \) is dimensionless.

4. **Final evaluation of the sine argument**:
- Since \( 2\pi \) is a constant (dimensionless), we have:
\[ \frac{2\pi}{\lambda}(vt - x) \text{ is dimensionless.} \]

### Step 3: Conclusion

Since the argument of the sine function is dimensionless, and both \( y \) and \( A \) have the dimension of length \([L]\), we can conclude that the left-hand side (LHS) and the right-hand side (RHS) of the equation have the same dimensions. Therefore, the equation is dimensionally correct.

### Final Result
The equation \( y = A \sin\left(\frac{2\pi}{\lambda}(vt - x)\right) \) is dimensionally correct.

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