In the expression ` y = a sin (omega t + theta )`, `y` is the displacement and `t` is the time . Write the dimension of` a `,` omega`and `theta` .

Explanation

To determine the dimensions of \( a \), \( \omega \), and \( \theta \) in the expression \( y = a \sin(\omega t + \theta) \), we will analyze each term step by step.

### Step 1: Identify the dimensions of \( y \)
In the given expression, \( y \) represents displacement. The dimension of displacement is the same as that of length. Therefore, we have:
\[ [y] = L \]

### Step 2: Determine the dimensions of \( a \)
Since \( \sin(\omega t + \theta) \) is a dimensionless quantity (as the sine function itself does not have dimensions), the term \( a \) must also have the same dimensions as \( y \) for the equation to be dimensionally consistent. Thus:
\[ [a] = [y] = L \]

### Step 3: Analyze the argument of the sine function
The argument of the sine function is \( \omega t + \theta \). For the sum \( \omega t + \theta \) to be valid, both terms must be dimensionless.

### Step 4: Determine the dimensions of \( \theta \)
Since \( \theta \) is an angle, it is a dimensionless quantity. We can express this as:
\[ [\theta] = M^0 L^0 T^0 \]

### Step 5: Determine the dimensions of \( \omega \)
Now, we need to analyze \( \omega t \). Since \( t \) represents time, its dimension is:
\[ [t] = T \]
For \( \omega t \) to be dimensionless, \( \omega \) must have dimensions that are the reciprocal of time. Therefore:
\[ [\omega] = T^{-1} \]

### Summary of Dimensions
- The dimension of \( a \) is \( L \) (length).
- The dimension of \( \omega \) is \( T^{-1} \) (reciprocal of time).
- The dimension of \( \theta \) is dimensionless, represented as \( M^0 L^0 T^0 \).

### Final Answer
- \( [a] = L \)
- \( [\omega] = T^{-1} \)
- \( [\theta] = M^0 L^0 T^0 \)