The dimensional formula for `ω` in the relation `y = A Sin ωt` is

Options

A. `[M° L° T]`
B. `[M° L° T^(–1)]`
C. `[ML° T°]`
D. `[M° L^(–1) T^(–1)]`

Explanation

To find the dimensional formula for \( \omega \) in the relation \( y = A \sin(\omega t) \), we can follow these steps:

### Step 1: Understand the equation
The equation \( y = A \sin(\omega t) \) involves the sine function, which requires its argument (in this case, \( \omega t \)) to be dimensionless. This means that the product \( \omega t \) must not have any dimensions.

### Step 2: Identify the dimensions of time
In this equation, \( t \) represents time. The dimensional formula for time is given as:
\[ [T] = T^1 \]

### Step 3: Set up the equation for \( \omega \)
Since \( \omega t \) must be dimensionless, we can express this as:
\[ [\omega t] = [1] \quad \text{(dimensionless)} \]
This implies:
\[ [\omega] \cdot [t] = 1 \]
Substituting the dimension of time:
\[ [\omega] \cdot T^1 = 1 \]

### Step 4: Solve for the dimensional formula of \( \omega \)
To make the left side dimensionless, we need:
\[ [\omega] = T^{-1} \]
This indicates that the dimensional formula for \( \omega \) is:
\[ [\omega] = T^{-1} \]

### Step 5: Write the complete dimensional formula
Since there are no mass or length components involved in \( \omega \), we can express the complete dimensional formula as:
\[ [\omega] = M^0 L^0 T^{-1} \]

### Final Answer
Thus, the dimensional formula for \( \omega \) in the relation \( y = A \sin(\omega t) \) is:
\[ M^0 L^0 T^{-1} \]

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