Q.The displacement of a progressive wave is represented by $y = A \sin(\omega t - k x)$, where $x$ is distance and $t$ is time. Write the dimensional formula of

Concept: Dimensional analysis of wave parameters

The argument of any trigonometric function must be dimensionless (a pure number). This constraint determines the dimensions of $\omega$ and $k$.

(i) Angular frequency $\omega$:

From $\sin(\omega t - kx)$, the term $\omega t$ must be dimensionless.

(ii) Wave number $k$: …

The sine function—like all trigonometric functions—accepts only dimensionless arguments. You cannot take the sine of "5 meters" or "3 seconds"; the input must be a pure number (whether in radians or degrees, both are dimensionless). This principle is the key to finding the dimensions of $\omega$ and $k$.

In the wave equation $y = A \sin(\omega t - kx)$, the entire argument $(\omega t - kx)$ must be dimensionless. For this to hold, each term inside must separately be dimensionless, because you can only add or subtract quantities with the same dimensions.

Finding the dimensional formula

1. Dimensional formula of $\omega$

The term $\omega t$ must be dimensionless. Since $t$ is time with dimension $[\mathrm{T}]$, we need:

where $[1]$ denotes a dimensionless quantity. Solving for $[\omega]$:

This makes physical sense: $\omega$ is the angular frequency, measured in radians per second (rad/s). Since radians are dimensionless, the dimension is simply inverse time.

2. Dimensional formula of $k$

Similarly, the term $kx$ must be dimensionless. Since $x$ is distance with dimension $[\mathrm{L}]$:

Solving for $[k]$:

$[k] = \frac{[1]}{[\mathrm{L}]} = [\mathrm{L}^{-1}]$ …

Concept: Dimensional Analysis of Wave Parameters

The argument of a sine function (or any trigonometric function) must be dimensionless — it is an angle (in radians), which has no dimensions.

Given:

$y = A \sin(\omega t - k x)$

Here:

• $t$ = time → dimension $[T]$
• $x$ = distance → dimension $[L]$

Method: Argument Dimensionless Principle

Step 1: The term $\omega t$ must be dimensionless.

Since $t$ has dimension $[T]$, we have:

Thus:

Step 2: The term $k x$ must also be dimensionless.

Since $x$ has dimension $[L]$, we have:

Thus:

--- …

Let’s break it down.

The Core Concept

In the wave equation

• The argument of a sine (or any trigonometric function) must be dimensionless.
• That means each term inside the sine — $\omega t$ and $k x$ — must individually have no dimensions.

So:

• $\omega t$ is dimensionless → tells us the dimensions of $\omega$
• $k x$ is dimensionless → tells us the dimensions of $k$

Common Mistake #1: Forgetting that the argument must be dimensionless

What students do wrong:

They directly write $\omega$ as having dimensions of $[T^{-1}]$ and $k$ as $[L^{-1}]$ without justification — or worse, they treat $\omega$ and $k$ as having the same dimensions as $t$ and $x$.

Why it’s wrong:

If $\omega$ had the same dimensions as $t$, then $\omega t$ would have dimensions of $[T^2]$, which is not allowed inside a sine.

How to avoid:

Always start with:

“Since $\omega t$ is dimensionless, $[\omega][T] = 1$”

So $[\omega] = [T^{-1}]$.

Similarly, $[k][L] = 1$ → $[k] = [L^{-1}]$.

Common Mistake #2: Confusing $\omega$ with angular frequency in circular motion

What students do wrong:

They remember $\omega = 2\pi f$ and write $[\omega] = [T^{-1}]$ — which is correct — but then they also write $[k] = [T^{-1}]$ or mix up the two.

Why it’s wrong:

$k$ is the wave number, not frequency. It relates to wavelength: $k = \frac{2\pi}{\lambda}$. So its dimension is inverse length, not inverse time.

How to avoid:

• $\omega$ → angular frequency → depends on time → $[T^{-1}]$
• $k$ → wave number → depends on space → $[L^{-1}]$

Memorise this pairing:

Time → $\omega$  Space → $k$

Common Mistake #3: Writing dimensions with incorrect brackets or missing the “inverse” sign

What students do wrong:

They write $[\omega] = T$ or $[k] = L$ (missing the negative exponent).

Why it’s wrong: …