Question

Force is given by the expression \( F = A \cos(Bx) + C \cos(Dt) \), where 'x' is displacement and 't' is time. The dimensions of \( \frac{D}{B} \) are the same as that of:

• A. Velocity
• B. Velocity gradient
• C. Angular velocity
• D. Angular momentum

Hint:

Understanding dimensional analysis can significantly aid in checking the consistency of equations and understanding the relationships between different physical quantities.

Answer 1

The Correct Option is A

Step 1: Understand the force expression. The force equation provided is: \[ F = A \cos(Bx) + C \cos(Dt) \] This suggests a dependence of force on both displacement \(x\) and time \(t\) through cosine functions. 
Step 2: Dimensional analysis of \( B \) and \( D \). \( B \) is associated with displacement \( x \) and thus has dimensions of \([L^{-1}]\) (inverse length). \( D \) is associated with time \( t \) and has dimensions of \([T^{-1}]\) (inverse time). 
Step 3: Analyze the dimensions of \( \frac{D}{B} \). Given \( D \) and \( B \) have dimensions: \[ [D] = T^{-1}, \quad [B] = L^{-1} \] The ratio \( \frac{D}{B} \) would then be: \[ \left[\frac{D}{B}\right] = \frac{T^{-1}}{L^{-1}} = \frac{L}{T} \] These are the dimensions of velocity, indicating that \( \frac{D}{B} \) is dimensionally equivalent to velocity. 
Step 4: Contextual understanding. Given that the force depends on both \( x \) and \( t \) through periodic functions, \( \frac{D}{B} \) comparing the rates of change with respect to space and time, directly corresponds to the velocity with which periodic patterns related to space and time offsets influence the force dynamics.