Question

When a wave traverses a medium the displacement of a particle located at $x$ at a time is given by $y = a \sin (bt - cx)$, where $a$, and $b$ are constants of the wave, which of the following is a quantity with dimensions?

• A. $\frac{y}{a} $
• B. $bt$
• C. $cx$
• D. $\frac{b}{c} $

Answer 1

The Correct Option is D

Given, \(y=a \sin (b t-c x)\) 
Comparing the given equation with general wave equation \(y=a \sin \frac{2 \pi t}{T}-\frac{2 \pi x}{\lambda}\) 
we get \(b=\frac{2 \pi}{T}, c=\frac{2 \pi}{\lambda}\) 
(a) Dimensions of \(\frac{y}{a}=\frac{\text { metre }}{\text { metre }}=\frac{L}{L}\) 
(b) Dimensions of \(b t=\frac{2 \pi}{T} . t=\frac{T}{T}\) 
(c) Dimensions of \(c x=\frac{2 \pi}{\lambda} \cdot x=\frac{L}{L}\) 
(d) Dimensions of \(\frac{b}{c}=\frac{2 \pi}{T} \frac{2 \pi}{\lambda}=\frac{\lambda}{T}=L T^{-1}\)

So, the correct option is (D): \(\frac bc\)

Answer 2

Given the wave equation \( y = a \sin (bt - cx) \), we need to determine which quantity has dimensions. Let's compare the given equation with the general wave equation:

\[ y = a \sin \left( \frac{2\pi t}{T} - \frac{2\pi x}{\lambda} \right) \]

From this comparison, we get:
- \( b = \frac{2\pi}{T} \)
- \( c = \frac{2\pi}{\lambda} \)

Let's analyze the dimensions of these terms:

(a) Dimensions of \(\frac{y}{a}\):

\[ \frac{y}{a} = \sin (bt - cx) \]

Since \( y \) (displacement) and \( a \) (amplitude) both have dimensions of length \([L]\):

\[ \frac{y}{a} = \frac{[L]}{[L]} = 1 \]
which is dimensionless.

(b) Dimensions of \( bt \):

\[ bt = \frac{2\pi}{T} \cdot t \]

where \( T \) is the time period and \( t \) is time. Thus,

\[ bt = \frac{[1]}{[T]} \cdot [T] = 1 \]
which is dimensionless.

(c) Dimensions of \( cx \):

\[ cx = \frac{2\pi}{\lambda} \cdot x \]

where \( \lambda \) is the wavelength and \( x \) is position. Thus,

\[ cx = \frac{[1]}{[L]} \cdot [L] = 1 \]
which is dimensionless.

(d) Dimensions of \( \frac{b}{c} \):

\[ \frac{b}{c} = \frac{\frac{2\pi}{T}}{\frac{2\pi}{\lambda}} = \frac{\lambda}{T} \]

\[ \frac{b}{c} = \frac{[L]}{[T]} = [L T^{-1}] \]

Thus, \( \frac{b}{c} \) has the dimensions of \([L T^{-1}]\), which is velocity.

Therefore, the quantity with dimensions is:

\[ \frac{b}{c} = \frac{\lambda}{T} = [L T^{-1}] \]

So, the correct answer is:

Option (d): \( \frac{b}{c} \) has dimensions.