The energy E of a particle varies with time t according to the equation `E=E_0sin(alphat).e^((-alphat)/(betax))`, where x is displacement from mean position `E_0` is energy at infinite position and `alpha` and `beta` are constants .
Dimensions of `sin((alphat)/(betax))` are

Options

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

Answer

To determine the dimensions of the expression \( \sin\left(\frac{\alpha t}{\beta x}\right) \), we need to ensure that the argument of the sine function is dimensionless. Here’s how we can approach the problem step by step:

### Step 1: Understand the Argument of the Sine Function
The argument of the sine function is given by \( \frac{\alpha t}{\beta x} \). For the sine function to be valid, its argument must be dimensionless.

### Step 2: Analyze the Components
- **Time (t)** has dimensions of \( [T] \).
- **Displacement (x)** has dimensions of \( [L] \).
- We need to determine the dimensions of \( \alpha \) and \( \beta \).

### Step 3: Determine the Dimensions of \( \alpha \) and \( \beta \)
1. **For \( \alpha \)**:
- Since \( \alpha t \) must have the same dimensions as \( \beta x \) for the fraction to be dimensionless, we can express this relationship.
- Let’s assume \( \alpha \) has dimensions \( [\alpha] \). Then, the dimensions of \( \alpha t \) are \( [\alpha][T] \).

2. **For \( \beta \)**:
- Similarly, let’s assume \( \beta \) has dimensions \( [\beta] \). Then, the dimensions of \( \beta x \) are \( [\beta][L] \).

### Step 4: Set Up the Dimensionless Condition
For \( \frac{\alpha t}{\beta x} \) to be dimensionless:
\[ [\alpha][T] = [\beta][L] \]
This implies:
\[ [\alpha] = \frac{[\beta][L]}{[T]} \]

### Step 5: Assign Dimensions to \( \alpha \) and \( \beta \)
To satisfy the dimensionless condition, we can assign:
- Let \( [\beta] = [L][T]^{-1} \) (which means \( \beta \) has dimensions of velocity).
- Then, substituting this into our previous equation gives:
\[ [\alpha] = \frac{[L][T]^{-1}[L]}{[T]} = [L^2][T]^{-2} \]

### Step 6: Conclusion on Dimensions
Since both \( \alpha t \) and \( \beta x \) have been shown to be dimensionally consistent, we conclude that:
\[ \frac{\alpha t}{\beta x} \text{ is dimensionless.} \]
Thus, the dimensions of \( \sin\left(\frac{\alpha t}{\beta x}\right) \) are dimensionless, which can be represented as \( [L^0 M^0 T^0] \).

### Final Answer
The dimensions of \( \sin\left(\frac{\alpha t}{\beta x}\right) \) are \( L^0 M^0 T^0 \) (dimensionless).

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