When a wave transverses in a medium, the displacement of a particle located at distance x at time t is given by `y=a sin (bt-cx)` where a, b and c are constants of the wave. The dimension of `b//c` are same as that of:

Options

A. wave velocity
B. wave length
C. wave amplitude
D. wave frequency

Answer

To solve the problem, we need to analyze the given wave equation and determine the dimensions of the constants \( b \) and \( c \).

### Step-by-Step Solution:

1. **Understand the Wave Equation**: The displacement of a particle in the wave is given by:
\[ y = a \sin(bt - cx) \]
Here, \( a \), \( b \), and \( c \) are constants.

2. **Identify the Argument of the Sine Function**: The argument of the sine function, \( (bt - cx) \), must be dimensionless. This means that both \( bt \) and \( cx \) must have no dimensions.

3. **Determine the Dimensions of \( b \)**:
- Since \( bt \) is dimensionless, we can express the dimensions of \( b \) as follows:
\[ [b][t] = 1 \quad \text{(dimensionless)} \]
- Rearranging gives:
\[ [b] = \frac{1}{[t]} \quad \text{(dimension of time)} \]
- Therefore, the dimension of \( b \) is:
\[ [b] = T^{-1} \]

4. **Determine the Dimensions of \( c \)**:
- Similarly, for \( cx \) to be dimensionless:
\[ [c][x] = 1 \quad \text{(dimensionless)} \]
- Rearranging gives:
\[ [c] = \frac{1}{[x]} \quad \text{(dimension of length)} \]
- Therefore, the dimension of \( c \) is:
\[ [c] = L^{-1} \]

5. **Calculate the Ratio \( \frac{b}{c} \)**:
- Now, we can find the dimensions of the ratio \( \frac{b}{c} \):
\[ \left[\frac{b}{c}\right] = \frac{[b]}{[c]} = \frac{T^{-1}}{L^{-1}} = \frac{1}{T} \cdot L = \frac{L}{T} \]
- The dimension \( \frac{L}{T} \) corresponds to velocity.

6. **Conclusion**:
- The dimensions of \( \frac{b}{c} \) are the same as that of wave velocity. Thus, the final answer is:
\[ \text{The dimensions of } \frac{b}{c} \text{ are the same as that of wave velocity.} \]