A travelling wave represented by y = A sin(wt - kx) is superimposed on another wave represented by y

Q: Why does the resultant equation represent a standing wave?

Because the spatial ($x$) and temporal ($t$) variables are separated into independent trigonometric functions. The wave doesn't propagate forward or backward; instead, it oscillates in place.

Q: What is the difference between a node and an antinode?

A node is a point on the standing wave that has zero amplitude (it never moves). An antinode is a point that oscillates with the maximum possible amplitude ($2A$).

Q: Why do we set $\cos kx = 0$ for nodes?

The amplitude of the standing wave at any point $x$ is given by the term $2A \cos kx$. Since nodes are defined as points with zero displacement at all times, their amplitude must be zero.

Q: How did we get $kx = (2n + 1)\frac{\pi}{2}$?

The cosine function equals zero at angles like $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}$, etc. These are all odd multiples of $\frac{\pi}{2}$, which can be written algebraically as $(2n + 1)\frac{\pi}{2}$ for integer values of $n$.

Q: What does $k$ represent in the wave equation?

$k$ is the angular wave number, defined as $k = \frac{2\pi}{\lambda}$. It relates the spatial frequency of the wave to its physical wavelength $\lambda$.

Question

A travelling wave represented by $y = A \sin(\omega t - kx)$ is superimposed on another wave represented by $y = A \sin(\omega t + kx)$. The resultant is

Options:

(A) a standing wave having nodes at $x = \left(n + \frac{1}{2}\right)\frac{\lambda}{2}, n = 0, 1, 2$

(B) a wave travelling along $+x$ direction

(C) a wave travelling along $-x$ direction

(D) a standing wave having nodes at $x = \frac{n\lambda}{2}; n = 0, 1, 2$

Sigma Percentile · Accepted Answer

### The Magic of Superposition

Imagine two identical waves traveling in opposite directions on a string. One wave is moving towards the right, represented by the equation $y_1 = A \sin(\omega t - kx)$, and the other is moving towards the left, represented by $y_2 = A \sin(\omega t + kx)$. What happens when these two waves meet?

According to the **Principle of Superposition**, the resultant displacement of the string at any point is simply the algebraic sum of the displacements caused by the individual waves. So, we write our master equation:

$$y = y_1 + y_2$$

### The Mathematical Symphony

Let's substitute our wave equations into the superposition principle:

$$y = A \sin(\omega t - kx) + A \sin(\omega t + kx)$$

To simplify this, we need a powerful tool from trigonometry: the sum-to-product formula. Recall that:

$$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$$

Let's assign $C = \omega t + kx$ and $D = \omega t - kx$. Plugging these into our identity, the magic happens. The $kx$ terms cancel out in the sine part, and the $\omega t$ terms cancel out in the cosine part. We are left with:

$$y = 2A \sin(\omega t) \cos(kx)$$

Rearranging this slightly gives us a beautiful insight:

$$y = (2A \cos kx) \sin(\omega t)$$

### Decoding the Standing Wave

Look closely at this final equation. Unlike a traveling wave where $x$ and $t$ are locked together in a phase term like $(\omega t - kx)$, here they are completely separated! The term $(2A \cos kx)$ acts as a spatial amplitude that depends *only* on position, while $\sin(\omega t)$ dictates the oscillation in time. This is the mathematical signature of a **standing wave**.

### Hunting for Nodes

In a standing wave, there are special points that never move. These points of zero amplitude are called **nodes**. To find them, we set our spatial amplitude to zero:

$$2A \cos kx = 0 \implies \cos kx = 0$$

We know from basic trigonometry that the cosine function is zero at odd multiples of $\frac{\pi}{2}$. Therefore:

$$kx = (2n + 1)\frac{\pi}{2}$$

where $n = 0, 1, 2, \dots$

Now, we recall the definition of the wave number, $k = \frac{2\pi}{\lambda}$. Substituting this into our equation:

$$\frac{2\pi}{\lambda}x = (2n + 1)\frac{\pi}{2}$$

Solving for $x$, we find the exact locations of the nodes:

$$x = (2n + 1)\frac{\lambda}{4} = \left(n + \frac{1}{2}\right)\frac{\lambda}{2}$$

This elegant mathematical derivation perfectly matches our physical intuition of standing waves, confirming that the nodes are spaced exactly half a wavelength apart!

Animated Solution for Physics - Waves: A travelling wave represented by $y = A sin (ω t - k x)$ is superimposed on another wave represented by $y = A sin (ω t + k x)$ . The resultant is

Select Answer:

Visualized Solution

Principle of Superposition

Trigonometric Identity

Applying the Identity

Resultant Wave Equation

Condition for Nodes

Solving for Node Positions

Final Answer

The Sigma Insight: Interference of Waves

The Magic of Superposition

The Mathematical Symphony

Decoding the Standing Wave

Hunting for Nodes

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