Standing waves can be produced
(a) on a string clamped at both ends
(b) on a string clamped at one e

Q: Why does a free end of a string reflect waves? I thought reflection only happens at rigid boundaries.

Reflection occurs whenever there is a change in the medium's properties (impedance mismatch). At a free end, the tension drops abruptly to zero, representing a boundary. The wave reflects back, but unlike a rigid boundary, it reflects in-phase (no phase change), creating an antinode.

Q: If the waves in option (d) have a phase difference of pi, why don't they form a standing wave?

For a standing wave, the waves must travel in opposite directions so that their spatial and temporal parts separate. Waves traveling in the same direction will always form a traveling wave (or cancel out completely if they are out of phase), but they can never create stationary nodes and antinodes.

Q: Do standing waves transmit energy?

No, ideal standing waves do not transmit energy. The energy is confined within the loops between the nodes. The energy continuously sloshes back and forth between purely kinetic energy (when the string is flat) and purely potential energy (when the string has maximum displacement).

Q: What happens if the reflecting wall in option (c) absorbs some energy?

If the wall absorbs energy, the reflected wave will have a smaller amplitude than the incident wave. In this case, we get a partial standing wave, which is a mixture of a standing wave and a traveling wave. The nodes will no longer have exactly zero amplitude.

Q: Are standing waves only transverse, or can they be longitudinal too?

Standing waves can be both! Transverse standing waves are formed on strings (like in a guitar), while longitudinal standing waves are formed by sound waves in air columns (like in organ pipes or flutes).

Question

Standing waves can be produced

Options:

(A) on a string clamped at both ends

(B) on a string clamped at one end and free at the other

(C) when incident wave gets reflected from a wall

(D) when two identical waves with a phase difference of $\pi$ are moving in the same direction

Sigma Percentile · Accepted Answer

### The Magic of Standing Waves

Imagine holding a skipping rope tied to a wall. If you wiggle your hand up and down, a wave travels down the rope, hits the wall, and bounces back.

If you time your wiggles just right, something magical happens: the rope stops looking like it is traveling. Instead, it seems to vibrate in place, forming beautiful, stationary loops.

This is the phenomenon of **standing waves** (or stationary waves). It is the physics behind the rich tones of a violin, the resonance of a flute, and even the quantum mechanical orbits of electrons in an atom!

Let's dive deep into the conditions required to create this beautiful dance of physics.

---

### The Core Principle: Superposition

To create a standing wave, we need two waves of the same frequency and amplitude traveling in **opposite directions** to superpose (interfere) with each other.

Let's write down the equations of these two waves mathematically:

$$y_1 = A \sin(kx - \omega t)$$

This represents a wave traveling in the positive $x$-direction.

$$y_2 = A \sin(kx + \omega t + \phi)$$

This represents an identical wave traveling in the negative $x$-direction, with a phase constant $\phi$.

When these two waves meet, they superpose. Using the trigonometric identity:

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

We get the resultant displacement:

$$y = y_1 + y_2 = 2A \sin\left(kx + \frac{\phi}{2}ight) \cos\left(\omega t + \frac{\phi}{2}ight)$$

Look closely at this resultant equation. The spatial part, $\sin\left(kx + \frac{\phi}{2}ight)$, is completely separated from the time-dependent part, $\cos\left(\omega t + \frac{\phi}{2}ight)$.

This separation of variables means that every particle of the medium vibrates in Simple Harmonic Motion (SHM) with a time-independent amplitude that depends only on its position $x$.

At certain points, called **nodes**, the amplitude is always zero. At other points, called **antinodes**, the amplitude is maximum ($2A$).

---

### Deconstructing the Options

Now, let's analyze each option of our problem to see if it satisfies this fundamental condition of counter-propagating waves.

#### Case 1: String Clamped at Both Ends

When a string is clamped tightly at both ends, any wave produced on it travels to one end, hits the rigid clamp, and undergoes reflection.

Because the clamp is a rigid boundary, the wave reflects with a phase change of $\pi$. This reflected wave travels in the opposite direction and superposes with the incident wave.

Since both ends are fixed, they must always remain stationary. This boundary condition requires nodes at both ends:

$$x = 0 \implies y = 0 \quad 	ext{and} \quad x = L \implies y = 0$$

This restricts the allowed wavelengths to discrete values:

$$\lambda_n = \frac{2L}{n} \quad (n = 1, 2, 3, \dots)$$

Thus, stable standing waves are easily produced. **Option (a) is correct.**

#### Case 2: String Clamped at One End and Free at the Other

What if one end of the string is free? Can a free end reflect waves?

Yes! When a wave reaches a free boundary, it reflects back without any phase change. This creates an antinode at the free end, while the clamped end remains a node.

Since we still have an incident wave going one way and a reflected wave returning in the opposite direction, they superpose to form standing waves. The allowed wavelengths are:

$$\lambda_n = \frac{4L}{2n-1} \quad (n = 1, 2, 3, \dots)$$

Thus, standing waves are produced here as well. **Option (b) is correct.**

#### Case 3: Reflection from a Wall

When an incident wave (like a sound wave or a light wave) strikes a solid wall, it gets reflected.

The incident wave and the reflected wave travel in opposite directions with the same frequency and amplitude. As they pass through each other, they continuously interfere, creating a stationary pattern of nodes and antinodes in front of the wall.

This is exactly how standing waves are formed in resonance columns and rooms. **Option (c) is correct.**

#### Case 4: Waves Moving in the Same Direction

What if the two waves are moving in the same direction, even with a phase difference of $\pi$?

Let's write their equations:

$$y_1 = A \sin(kx - \omega t)$$

$$y_2 = A \sin(kx - \omega t + \pi) = -A \sin(kx - \omega t)$$

Superposing them yields:

$$y = y_1 + y_2 = 0$$

They completely cancel each other out (destructive interference). If the phase difference were anything other than $\pi$, say $\phi$, they would combine to form another traveling wave:

$$y = 2A \cos\left(\frac{\phi}{2}ight) \sin\left(kx - \omega t + \frac{\phi}{2}ight)$$

Since the spatial and temporal parts are not separated, this is a traveling wave, not a standing wave. **Option (d) is incorrect.**

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### Conclusion

Standing waves are a beautiful consequence of boundary conditions and wave reflection. They require waves traveling in opposite directions, which is naturally achieved through reflection at boundaries (clamped or free) or off walls.

Therefore, the correct options are **(a), (b), and (c)**.

Animated Solution for Physics - Waves: Standing waves can be produced

Select Answer:

Visualized Solution

Understanding Standing Waves

Mathematical Condition for Standing Waves

String Clamped at Both Ends

String Clamped at One End and Free at the Other

Reflection from a Wall

Waves Moving in the Same Direction

Summary of Correct Options

Exploring Further

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Magic of Standing Waves

The Core Principle

Deconstructing the Options

# Case 1

# Case 2

# Case 3

# Case 4

Conclusion

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