Question

In the travelling plane wave equation given by \( y = A \sin \omega \left( \frac{x}{v} - t \right) \), where \( \omega \) is the angular velocity and \( v \) is the linear velocity. 
The dimension of \( \omega t \) is:

• A. \( LM^\circ{T}^{-1} \)
• B. \( {L}^\circ{M}^\circ{T}^\circ \)
• C. \( {L}^\circ{M}^\circ{T} \)
• D. \( {LMT} \)
• E. \( {LMT}^{-2} \)

Hint:

For dimensionless quantities like \( \omega t \), the total dimensional contribution becomes zero, as they do not contribute any physical dimension.

Answer 1

The Correct Option is B

We are given the equation for the travelling wave as: \[ y = A \sin \omega \left( \frac{x}{v} - t \right) \] where \( y \) is the displacement, \( A \) is the amplitude, \( \omega \) is the angular velocity, \( v \) is the linear velocity, \( x \) is the position, and \( t \) is time. 
We need to find the dimensions of \( \omega t \). 
Step 1: The general dimension formula for angular velocity \( \omega \) is: \[ {Dimension of } \omega = \frac{{Dimension of angular displacement}}{{Dimension of time}} = \frac{{L}^0 {M}^0 {T}^0}{{T}} = {T}^{-1}. \] Step 2: The dimension of time \( t \) is simply: \[ {Dimension of } t = {T}. \] Step 3: Thus, the product \( \omega t \) has the dimension: \[ {Dimension of } \omega t = {T}^{-1} \times {T} = {T}^0. \] Therefore, the dimension of \( \omega t \) is \( {L}^0 {M}^0 {T}^0 \).