The equation of a wave is given by `Y = A sin omega((x)/(v) -k)`
where `omega` is the angular velocity and v is the linear velocity. The dimension of k is

Options

A. `[LT]`
B. `[T]`
C. `[T^(-1)]`
D. `[T^(2)]`

Explanation

To find the dimension of \( k \) in the wave equation \( Y = A \sin \left( \omega \left( \frac{x}{v} \right) - k \right) \), we need to analyze the terms inside the sine function.

### Step 1: Understand the terms in the equation
The argument of the sine function must be dimensionless. Therefore, the term \( \omega \left( \frac{x}{v} \right) - k \) must also be dimensionless.

### Step 2: Analyze \( \frac{x}{v} \)
Here, \( x \) represents distance, and its dimension is:
\[ [x] = L \quad (\text{where } L \text{ is length}) \]
The linear velocity \( v \) has the dimension:
\[ [v] = \frac{L}{T} \quad (\text{where } T \text{ is time}) \]
Now, we can find the dimension of \( \frac{x}{v} \):
\[ \left[\frac{x}{v}\right] = \frac{[x]}{[v]} = \frac{L}{\frac{L}{T}} = T \]
Thus, the dimension of \( \frac{x}{v} \) is \( T \).

### Step 3: Analyze \( \omega \)
The angular velocity \( \omega \) has the dimension:
\[ [\omega] = \frac{1}{T} \]
This is because angular velocity is defined as the angle (in radians) per unit time.

### Step 4: Combine \( \omega \) and \( \frac{x}{v} \)
Now, we can analyze the term \( \omega \left( \frac{x}{v} \right) \):
\[ \left[\omega \left( \frac{x}{v} \right)\right] = \left[\omega\right] \cdot \left[\frac{x}{v}\right] = \frac{1}{T} \cdot T = 1 \quad (\text{dimensionless}) \]
This confirms that \( \omega \left( \frac{x}{v} \right) \) is dimensionless.

### Step 5: Determine the dimension of \( k \)
Since the entire expression \( \omega \left( \frac{x}{v} \right) - k \) must be dimensionless, and we have established that \( \omega \left( \frac{x}{v} \right) \) is dimensionless, it follows that \( k \) must also be dimensionless:
\[ [k] = 1 \quad (\text{dimensionless}) \]

### Conclusion
The dimension of \( k \) is dimensionless.

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