In the relation `( dy)/( dt) = 2 omega sin ( omega t + phi_(0))`, the dimensional formula for ` omega t + phi_(0)` is

Options

A. `MLT`
B. `MLT^(0)`
C. `ML^(0) T^(0)`
D. `M^(0) L^(0) T^(0)`

Explanation

To find the dimensional formula for the expression \( \omega t + \phi_0 \) in the relation

\[ \frac{dy}{dt} = 2 \omega \sin(\omega t + \phi_0), \]

we need to analyze the components of this expression step by step.

### Step 1: Understand the components of the expression

The expression \( \omega t + \phi_0 \) consists of two parts: \( \omega t \) and \( \phi_0 \).

### Step 2: Analyze \( \omega t \)

- Here, \( \omega \) represents angular frequency, which is defined as the rate of change of the angle with respect to time. The dimensional formula for angular frequency \( \omega \) is given by:

\[ \omega = \frac{2\pi}{T} \quad \text{(where \( T \) is the time period)} \]

Thus, the dimensional formula for \( \omega \) is:

\[ [\omega] = [T^{-1}] \]

- The variable \( t \) represents time, and its dimensional formula is:

\[ [t] = [T] \]

- Now, when we multiply \( \omega \) and \( t \):

\[ [\omega t] = [T^{-1}] \cdot [T] = [T^0] = [1] \quad \text{(dimensionless)} \]

### Step 3: Analyze \( \phi_0 \)

- The term \( \phi_0 \) represents a phase constant, which is also a measure of angle. Angles are typically measured in radians or degrees, both of which are dimensionless quantities.

Thus, the dimensional formula for \( \phi_0 \) is:

\[ [\phi_0] = [1] \quad \text{(dimensionless)} \]

### Step 4: Combine the components

Now we can combine the two parts of the expression:

\[ \omega t + \phi_0 \]

Since both \( \omega t \) and \( \phi_0 \) are dimensionless, their sum is also dimensionless:

\[ [\omega t + \phi_0] = [1] + [1] = [1] \quad \text{(dimensionless)} \]

### Conclusion

The dimensional formula for \( \omega t + \phi_0 \) is:

\[ [M^0 L^0 T^0] \quad \text{or simply } [1] \]

### Final Answer

The dimensional formula for \( \omega t + \phi_0 \) is \( M^0 L^0 T^0 \) (dimensionless).

---