Question

The equation of stationary wave is given as \(y=2A\sin(\frac{2\pi}{\lambda}\cdot nt) \cos (\frac{2\pi}{\lambda}\cdot x)\), then which of the following is not correct.

• A. Dimension of x is [L]
• B. Dimension of n is [LT^-1]
• C. Dimension of \(\frac{n}{\lambda}\) is [T]
• D. Dimension of nt is [L]

Answer 1

The Correct Option is C

To solve the given problem, we need to analyze each option in terms of dimensional analysis. Dimensional analysis is a process that checks the consistency of equations by comparing the dimensions on both sides.

The equation of the stationary wave mentioned is:

\(y = 2A \sin\left(\frac{2\pi}{\lambda} \cdot nt\right) \cos\left(\frac{2\pi}{\lambda} \cdot x\right)\) 

Let's analyze each option:

1. Dimension of \(x\) is [L]:
   • In the term \(\cos\left(\frac{2\pi}{\lambda} \cdot x\right)\), the argument of a cosine function is dimensionless.
   • This implies that \(x\) divided by \(\lambda\) (a length) is dimensionless, i.e., \([\frac{x}{\lambda}] = [1]\).
   • Thus, the dimension of \(x\) is indeed [L].
2. Dimension of \(n\) is [LT^-1]:
   • In the term \(\sin\left(\frac{2\pi}{\lambda} \cdot nt \right)\), for the argument to be dimensionless, \(\frac{nt}{\lambda}\) should be dimensionless.
   • Thus, \([nt] = [L]\), where \(\lambda\) has a dimension [L]. Hence, \([n] = [L][T^{-1}]\).
3. Dimension of \(\frac{n}{\lambda}\) is [T]:
   • The dimension of \(n\) is found to be [LT^-1].
   • The dimension of \(\lambda\) is [L].
   • Thus, \(\frac{n}{\lambda} = \frac{[L T^{-1}]}{[L]} = [T^{-1}]\), not [T].
   • Therefore, this option is incorrect according to dimensional analysis.
4. Dimension of \(nt\) is [L]:
   • As detailed in option 2, for the argument to be dimensionless: \(\frac{nt}{\lambda} = 1\).
   • This implies that \([nt] = [L]\), as \(\lambda\) is [L].

Based on this analysis, the incorrect statement with respect to dimensions is:

The dimension of \(\frac{n}{\lambda}\) is [T].