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In the relation ( dy)/( dt) = 2 omega si...

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

A

`[MLT]`

B

`[MLT^0]`

C

`[ML^0T0]`

D

`[M^0L^0T^0]`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the expression given in the question: \[ \frac{dy}{dt} = 2 \omega \sin(\omega t + \phi_0) \] ### Step 1: Understand the components of the equation - Here, \( \omega \) is the angular frequency, which has dimensions of \( T^{-1} \) (inverse of time). - \( t \) is time, which has dimensions of \( T \). - \( \phi_0 \) is a phase constant, which is also dimensionless. ### Step 2: Analyze the argument of the sine function The argument of the sine function is \( \omega t + \phi_0 \). We need to determine the dimensions of this entire expression. ### Step 3: Calculate the dimensions of \( \omega t \) - Since \( \omega \) has dimensions of \( T^{-1} \) and \( t \) has dimensions of \( T \), we can calculate the dimensions of \( \omega t \): \[ [\omega t] = [\omega][t] = T^{-1} \cdot T = 1 \quad \text{(dimensionless)} \] ### Step 4: Analyze the phase constant \( \phi_0 \) - The phase constant \( \phi_0 \) is also dimensionless. Therefore, it does not contribute any dimensions. ### Step 5: Combine the dimensions - Since both \( \omega t \) and \( \phi_0 \) are dimensionless, their sum \( \omega t + \phi_0 \) is also dimensionless. ### Conclusion The dimensional formula for \( \omega t + \phi_0 \) is: \[ M^0 L^0 T^0 \quad \text{(dimensionless)} \] Thus, the answer is \( M^0 L^0 T^0 \). ---
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