In equation `varepsilon =A sin (omegat-kx)` where t is in second and x is in meter, the dimensional formula for

Options

A. `omega "is" [M^0L^0T^(-1)]`
B. k is `[M^0L^(-1)T^(0)]`
C. `(omega)/k "is" [M^0LT^(-1)]`
D. A is `[M^0LT^(-1)]`

Answer

To solve the problem, we need to find the dimensional formulas for the quantities involved in the equation \( \varepsilon = A \sin(\omega t - kx) \). Here, \( t \) is in seconds and \( x \) is in meters. We will analyze each component step by step.

### Step 1: Finding the dimensional formula for \( \omega \)

1. **Understanding the equation**: The argument of the sine function \( \omega t - kx \) must be dimensionless. This means that the dimensions of \( \omega t \) must equal the dimensions of \( kx \).

2. **Setting up the relationship**: Since both terms must be dimensionless, we can write:
\[ [\omega t] = [kx] \]

3. **Expressing \( \omega \)**: Rearranging gives us:
\[ \omega = \frac{\theta}{t} \]
where \( \theta \) is dimensionless.

4. **Finding the dimensions**: Since \( t \) has dimensions of time:
\[ [\omega] = \frac{1}{[t]} = [t]^{-1} \]
Thus, the dimensional formula for \( \omega \) is:
\[ [\omega] = M^0 L^0 T^{-1} \]

### Step 2: Finding the dimensional formula for \( k \)

1. **Using the relationship**: Similarly, we can express \( k \) using the relationship:
\[ k = \frac{\theta}{x} \]

2. **Finding the dimensions**: Since \( x \) has dimensions of length:
\[ [k] = \frac{1}{[x]} = [L]^{-1} \]
Thus, the dimensional formula for \( k \) is:
\[ [k] = M^0 L^{-1} T^0 \]

### Step 3: Finding the dimensional formula for \( \frac{\omega}{k} \)

1. **Dividing the dimensions**: We can find the dimensions of \( \frac{\omega}{k} \) by dividing the dimensional formulas of \( \omega \) and \( k \):
\[ \left[\frac{\omega}{k}\right] = \frac{[M^0 L^0 T^{-1}]}{[M^0 L^{-1} T^0]} = M^{0} L^{1} T^{-1} \]

### Step 4: Finding the dimensional formula for \( A \)

1. **Understanding amplitude**: The amplitude \( A \) represents the maximum displacement in a wave.

2. **Finding the dimensions**: The dimensional formula for displacement is:
\[ [A] = [L] \]
Thus, the dimensional formula for \( A \) is:
\[ [A] = M^0 L^{1} T^0 \]

### Summary of Dimensional Formulas

- \( [\omega] = M^0 L^0 T^{-1} \)
- \( [k] = M^0 L^{-1} T^0 \)
- \( \left[\frac{\omega}{k}\right] = M^0 L^{1} T^{-1} \)
- \( [A] = M^0 L^{1} T^0 \)