Question

In the relation \(y = r\sin(\omega t - kx)\), the dimensional formula of \(\dfrac{\omega}{k}\) is:

• A. \([M^0 L^0 T^0]\)
• B. \([M^0 L^1 T^{-1}]\)
• C. \([M^0 L^0 T^{-1}]\)
• D. \([M^0 L^1 T^0]\)

Hint:

In wave equations:
\(\omega\) has dimension \(T^{-1}\)
\(k\) has dimension \(L^{-1}\)
\(\dfrac{\omega}{k}\) always gives wave speed Use dimensional consistency to identify physical meaning.

Answer 1

The Correct Option is B

Step 1: Use the condition for arguments of trigonometric functions. The argument of the sine function must be dimensionless: \[ \omega t - kx \;\; \text{is dimensionless} \]
Step 2: Find the dimensions of \(\omega\) and \(k\). From \(\omega t\) being dimensionless: \[ [\omega][t] = 1 \Rightarrow [\omega] = T^{-1} \] From \(kx\) being dimensionless: \[ [k][x] = 1 \Rightarrow [k] = L^{-1} \]
Step 3: Find the dimensional formula of \(\dfrac{\omega}{k}\). \[ \left[\frac{\omega}{k}\right] = \frac{T^{-1}}{L^{-1}} = LT^{-1} \]
Step 4: Write in standard dimensional form. \[ \boxed{[M^0 L^1 T^{-1}]} \] Physical Interpretation: \[ \frac{\omega}{k} = v \] which represents the wave velocity.