In the relation ` y = r sin ( omega t - kx)`, the dimensions of `omega//k` are

Options

A. `[M^(0) L^(0) T^(0)]`
B. `[M^(0) L^(1) T^(-1)]`
C. `[M^(0) L^(0) T^(1)]`
D. `[M^(0) L^(1) T^(0)]`

Explanation

To find the dimensions of the ratio \(\frac{\omega}{k}\) from the relation \(y = r \sin(\omega t - kx)\), we will follow these steps:

### Step 1: Understand the terms in the equation
In the equation \(y = r \sin(\omega t - kx)\):
- \(y\) is a displacement (length), so its dimension is \([L]\).
- \(r\) is a constant (amplitude), also with dimension \([L]\).
- \(\omega\) is the angular frequency, and \(k\) is the wave number.

### Step 2: Determine the dimensions of \(\omega\)
The term \(\omega t\) must be dimensionless since it is inside the sine function. Therefore:
- Let the dimension of \(\omega\) be \([\omega]\).
- The dimension of time \(t\) is \([T]\).

Since \(\omega t\) is dimensionless:
\[ [\omega] \cdot [T] = 1 \implies [\omega] = [T]^{-1} = T^{-1} \]

### Step 3: Determine the dimensions of \(k\)
The term \(kx\) must also be dimensionless. Therefore:
- Let the dimension of \(k\) be \([k]\).
- The dimension of position \(x\) is \([L]\).

Since \(kx\) is dimensionless:
\[ [k] \cdot [L] = 1 \implies [k] = [L]^{-1} = L^{-1} \]

### Step 4: Find the dimensions of \(\frac{\omega}{k}\)
Now we can find the dimensions of the ratio \(\frac{\omega}{k}\):
\[ \frac{\omega}{k} = \frac{[T]^{-1}}{[L]^{-1}} = [T]^{-1} \cdot [L]^{1} = \frac{L}{T} \]

### Conclusion
The dimensions of \(\frac{\omega}{k}\) are \([L][T]^{-1}\), which is the dimension of velocity.

### Final Answer
\[ \text{Dimensions of } \frac{\omega}{k} = \frac{L}{T} \]

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