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 find the dimensions of \( a \), \( \omega \), and \( \theta \) in the expression \( y = a \sin(\omega t + \theta) \), we can follow these steps:

### Step 1: Understand the Variables
- \( y \) is the displacement, which has the dimension of length. Therefore, the dimension of \( y \) is \( [y] = L^1 \).
- \( t \) is time, which has the dimension of time. Therefore, the dimension of \( t \) is \( [t] = T^1 \).

### Step 2: Analyze the Sine Function
The sine function, \( \sin(x) \), is dimensionless. This means that the argument of the sine function, \( \omega t + \theta \), must also be dimensionless.

### Step 3: Determine the Dimension of \( \omega \)
Since \( \omega t \) must be dimensionless, we can express this as:
\[ [\omega t] = [\omega][t] = [\omega][T^1] \]
To make this dimensionless, we need:
\[ [\omega] [T^1] = L^0 M^0 T^0 \]
This implies:
\[ [\omega] = T^{-1} \]

### Step 4: Determine the Dimension of \( \theta \)
Since \( \theta \) is added to \( \omega t \) in the sine function, it must also be dimensionless. Therefore:
\[ [\theta] = L^0 M^0 T^0 \]
This means that \( \theta \) is dimensionless.

### Step 5: Determine the Dimension of \( a \)
From the equation \( y = a \sin(\omega t + \theta) \), since \( \sin(\omega t + \theta) \) is dimensionless, the dimensions of \( a \) must match those of \( y \):
\[ [a] = [y] = L^1 \]

### Summary of Dimensions
- The dimension of \( a \) is \( [a] = L^1 \).
- The dimension of \( \omega \) is \( [\omega] = T^{-1} \).
- The dimension of \( \theta \) is \( [\theta] = L^0 M^0 T^0 \) (dimensionless).

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

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