Problem 5 - Master Resource Book in JEE Main Physics

Question

When a wave transverses a medium the displacement of a particle located at \(x\) at a time \(t\) is given by \(y=a \sin (b t-c x)\), where \(a, b\) and \(c\) are constants of the wave. Which of the following is dimensionless? \(\begin{array}{llll}\text { (a) } \frac{y}{a} & \text { (b) } b t & \text { (c) } c x & \text { (d) } \frac{b}{c}\end{array}\)

Step-by-Step Solution

Answer

(a), (b), and (c) are dimensionless.

Step 1: Understanding dimensional analysis

Dimensional analysis involves comparing the dimensions of physical quantities to determine if one quantity divided by another is dimensionless. A dimensionless quantity has no units or dimensions and is often the result of a ratio.

Step 2: Analyzing choice (a)

The expression \( \frac{y}{a} \) simplifies the displacement \( y \) by \( a \). Since \( a \) is the amplitude (having dimensions of length) and \( y \) is also a displacement (length), their ratio \( \frac{y}{a} \) is dimensionless because the length dimensions cancel each other out.

Step 3: Analyzing choice (b)

For \( b t \), we consider \( b \) and \( t \). If \( b \) has dimensions related to frequency (\( T^{-1} \)), and \( t \) is time (\( T \)), \( b t \) is dimensionless because the time dimensions cancel, \( T \times T^{-1} = 1 \).

Step 4: Analyzing choice (c)

For \( c x \), if \( c \) has units of \( L^{-1} \) (inverse length, related to spatial frequency), and \( x \) is a distance (\( L \)), then \( c x \) is dimensionless, as the length dimensions cancel out, \( L \times L^{-1} = 1 \).

Step 5: Analyzing choice (d)

For \( \frac{b}{c} \), we compare the dimensions of \( b \) and \( c \). Since \( b \) has dimensions of frequency (\( T^{-1} \)), and \( c \) is spatial frequency (\( L^{-1} \)), \( \frac{b}{c} \) has the dimensions \( \frac{T^{-1}}{L^{-1}} = \frac{L}{T} \), which is not dimensionless.

Step 6: Final determination of dimensionless quantities

After analyzing all options, we can conclude that choices (a), (b), and (c) give dimensionless quantities: \( \frac{y}{a} \), \( b t \), and \( c x \).

Key Concepts

Wave Equation

The wave equation is a fundamental representation used to describe the motion of waves through a medium. In this context, the wave equation is given by \( y = a \sin(bt - cx) \), where:

• \( y \) is the displacement of the wave.
• \( a \) is the amplitude, representing the maximum displacement from the equilibrium point.
• \( b \) and \( c \) are constants related to the frequency and spatial characteristics of the wave.
• \( x \) and \( t \) denote position and time, respectively.

The sine function models the oscillatory nature of the wave, influenced by time and spatial variables. Waves are essentially disturbances that travel through a medium. In this equation, such a disturbance is captured by the sinusoidal variation, characterized by periodicity and locations.
Understanding this equation lays the groundwork for grasping how waves behave under different conditions, such as varying frequencies or mediums. Recognizing that the terms inside the sine function must be dimensionless is central to solving related problems.

Dimensional Quantities

Dimensional analysis is a powerful tool in physics that helps us understand the nature of physical quantities by breaking them down into their constituent dimensions such as length \( (L) \), time \( (T) \), and mass \( (M) \). In the context of the given wave equation, the dimensional analysis is applied to identify which components or ratios are dimensionless.

• Choice (a), \( \frac{y}{a} \), involves both \( y \) and \( a \), which have dimensions of length \( (L) \). Thus, their ratio is dimensionless as the dimensions cancel out.
• Choice (b), \( bt \), deals with time. If \( b \) has dimensions of frequency \( (T^{-1}) \) and \( t \) represents time \( (T) \), then the product \( bt \) is dimensionless as \( T \times T^{-1} = 1 \).
• Choice (c), \( cx \), uses spatial frequency. Considering \( c \) with dimensions of \( (L^{-1}) \) and \( x \) as length \( (L) \), their product results in a dimensionless quantity, \( L \times L^{-1} = 1 \).
• Choice (d), \( \frac{b}{c} \), is not dimensionless as it combines differing dimensions: frequency to spatial frequency, \( \frac{T^{-1}}{L^{-1}} = \frac{L}{T} \).

Dimensional analysis helps validate the equations and assumptions by ensuring that the physical relations maintain their integrity in terms of units and dimensions, crucial for understanding physics at a deeper level.

Frequency Analysis

Frequency analysis in the context of wave phenomena refers to examining how often the periodic processes repeat over time. The wave function \( y = a \sin(bt - cx) \) includes factors associated with both time-based frequency and spatial frequency.

• The term \( b \) relates to the temporal frequency, determining how rapidly the wave oscillates over time. Its dimension is \( T^{-1} \), indicating occurrences per unit time.
• The term \( c \) is linked to spatial frequency, often referred to as wavenumber, indicating how many cycles occur per unit distance, with dimensions of \( L^{-1} \).

By examining the time component, \( bt \), and the spatial component, \( cx \), we can understand how wave properties change over time and across different locations within a medium. In dimensional analysis, evaluating the frequency dimensions also helps determine if their combination with other units, such as time or length, results in a dimensionless term. Such analysis is crucial when solving differential equations that describe waves, ensuring that expressions make physical sense and coincide with the natural world.