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 .
Dimensional formula of `alpha` is

Options

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

Answer

To find the dimensional formula of the constant \(\alpha\) in the given equation \(E = E_0 \sin(\alpha t) \cdot e^{-\frac{\alpha t}{\beta x}}\), we will follow these steps:

### Step 1: Identify the terms in the equation
The equation involves:
- \(E\): Energy
- \(E_0\): Energy at infinite position
- \(\alpha\): A constant that we need to find the dimensions of
- \(t\): Time
- \(x\): Displacement from the mean position

### Step 2: Analyze the sine function
The term \(\sin(\alpha t)\) must be dimensionless because the sine function can only take dimensionless arguments. This means that the product \(\alpha t\) must also be dimensionless.

### Step 3: Write the dimensional formula for time
The dimensional formula for time \(t\) is given as:
\[ [T] = T^1 \]

### Step 4: Set up the equation for \(\alpha\)
Since \(\alpha t\) is dimensionless, we can express this as:
\[ [\alpha t] = [L^0 M^0 T^0] \]
This implies:
\[ [\alpha] \cdot [T^1] = [L^0 M^0 T^0] \]

### Step 5: Solve for the dimensions of \(\alpha\)
To isolate \([\alpha]\), we rearrange the equation:
\[ [\alpha] = \frac{[L^0 M^0 T^0]}{[T^1]} = [L^0 M^0 T^{-1}] \]

### Step 6: Write the final dimensional formula
Thus, the dimensional formula of \(\alpha\) is:
\[ [L^0 M^0 T^{-1}] \quad \text{or simply} \quad T^{-1} \]

### Conclusion
The dimensional formula for \(\alpha\) is:
\[ [L^0 M^0 T^{-1}] \]