Question

y = a sin(βx + γt)wherex and t represent displacement and time, respectively. Then, the dimensional formula for β— γis:

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

Hint:

In wave equations of the form \( y = a \sin (\beta x + \gamma t) \), - \( \beta \) (wave number) has a dimension of \( [L^{-1}] \). - \( \gamma \) (angular frequency coefficient) has a dimension of \( [T^{-1}] \). - The ratio \( \frac{\beta}{\gamma} \) represents wave velocity, which has a dimension of \( [M^0L^1T^{-1}] \).

Answer 1

The Correct Option is A

Step 1: Understanding the given equation The given equation represents a wave motion: \[ y = a \sin (\beta x + \gamma t) \] where: - \( x \) is displacement (dimension: \([L]\)) - \( t \) is time (dimension: \([T]\)) - \( \beta \) and \( \gamma \) are coefficients corresponding to spatial and temporal frequencies, respectively. 

Step 2: Finding dimensions of \( \beta \) and \( \gamma \) Since the argument of sine function must be dimensionless, \[ \beta x \quad \text{(must be dimensionless)} \Rightarrow \beta = \frac{1}{x} \] Thus, the dimension of \( \beta \) is: \[ [\beta] = [L^{-1}] \] Similarly, \[ \gamma t \quad \text{(must be dimensionless)} \Rightarrow \gamma = \frac{1}{t} \] Thus, the dimension of \( \gamma \) is: \[ [\gamma] = [T^{-1}] \] 

Step 3: Finding the dimension of \( \frac{\beta}{\gamma} \) \[ \frac{\beta}{\gamma} = \frac{[L^{-1}]}{[T^{-1}]} \] \[ = [L^{-1} T^{1}] \] Rearranging, \[ = [M^0 L^1 T^{-1}] \] 

Step 4: Verifying the correct option Comparing with given options, the correct answer is: \[ \mathbf{[M^0L^1T^{-1}]} \]