Wave Mechanics Examples: Standing Waves

This companion article provides exercises for WM10, wave mechanics: Standing Waves. The exercises are stated first so they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

The central standing-wave result from WM10 is

(1)

for the superposition of two equal sinusoidal waves traveling in opposite directions,

u1(x,t)	= A cos(kx − ωt),	(2)
u2(x,t)	= A cos(kx + ωt).	(3)

For this cosine-form standing wave,

(4)

and the invariant spacing rules are

(5)

The local amplitude magnitude is

(6)

In the later part of the standing-wave story, boundary conditions pick out the allowed normal modes. For a string fixed at both ends,

(7)

so that

(8)

How to use this problem set

Attempt all exercises in Part I before consulting Part II. For each problem, be explicit about whether you are identifying a node, an antinode, a wavelength, a local amplitude, a temporal phase relation, or a boundary-selected mode. These ideas are closely related but not interchangeable.

Part I: Exercises

Exercise 1: Derive the standing-wave form

Starting from

u1(x,t)	= A cos(kx − ωt),	(9)
u2(x,t)	= A cos(kx + ωt),	(10)

use the identity

(11)

to show that their superposition is a standing wave.

Then answer:

1. Write the final expression for u(x,t).
2. Which factor depends only on position?
3. Which factor depends only on time?
4. Why is the result not a single rigidly translating wave of the form F(x ∓ ct)?

Exercise 2: Read node and antinode structure from a snapshot

The following figure shows one standing-wave snapshot, with nodes and antinodes marked.

Use the figure to answer:

1. If the distance from N₁ to N₂ is 0.24 m, what is the wavelength?
2. What is the distance from N₂ to the nearest antinode?
3. If the antinode at A₂ is located at x = 0.48 m, where are the neighboring nodes?
4. Explain why a snapshot alone does not tell you whether the wave is standing or traveling unless you also know the time behavior.

Exercise 3: Locate nodes and antinodes from an equation

A standing wave is given by

(12)

where x is in meters and t is in seconds.

1. Identify the coefficient k.
2. Find the wavelength.
3. Find the first three nonnegative antinode positions.
4. Find the first three positive node positions.
5. What is the maximum displacement magnitude at an antinode?

Exercise 4: Local amplitude at specific positions

For the same standing wave as in Exercise 3,

(13)

find the local amplitude magnitude at the following positions:

1. x = 0,
2. x = 0.10 m,
3. x = 0.20 m,
4. x = 0.30 m.

At which of these positions is the point a node? At which is it an antinode?

Exercise 5: Relative temporal phase of different positions

The next figure marks three positions P, Q, and R on a standing wave.

Answer the following:

1. Are P and Q in the same or opposite temporal phase?
2. Are P and R in the same or opposite temporal phase?
3. Which point has the largest local amplitude?
4. Explain why crossing a node changes the temporal phase relation by π.

Exercise 6: Recover the component traveling waves

A standing wave is written as

(14)

1. What is the amplitude of each component traveling wave?
2. Write one possible right-moving component.
3. Write one possible left-moving component.
4. If the standing-wave maximum displacement is doubled, by what factor do the component-wave amplitudes change?

Exercise 7: Determine the times of special motion

A point is located at an antinode of the standing wave

(15)

1. What is the angular frequency ω?
2. What is the ordinary frequency f?
3. At what times in the interval 0 ≤ t ≤ 0.08 s does this antinode pass through zero displacement?
4. At what times in the same interval does it reach maximum positive displacement?
5. At what times in the same interval does it reach maximum negative displacement?

Exercise 8: Standing waves on a string with fixed ends

A string has length

(16)

and both ends are fixed.

1. What wavelengths are allowed for the first three normal modes?
2. How many interior nodes occur in the third mode?
3. What is the separation between adjacent nodes in the second mode?
4. What is the separation between adjacent antinodes in the third mode?

The figure below shows the first three fixed-end modes.

Exercise 9: Mode frequencies from wave speed

For the same string in Exercise 8, suppose the wave speed is

(17)

Find:

1. the fundamental frequency f₁,
2. the second-harmonic frequency f₂,
3. the third-harmonic frequency f₃,
4. the ratio f₃∕f₁.

Exercise 10: Diagnose conceptual statements

For each statement, decide whether it is correct. If it is incorrect, rewrite it accurately.

1. “A standing wave means every point of the medium is motionless.”
2. “Adjacent nodes are one full wavelength apart.”
3. “All points between two neighboring nodes have the same amplitude.”
4. “Points in adjacent lobes move in opposite temporal phase.”
5. “Every superposition of two opposite-going waves produces a perfect standing wave.”

Exercise 11: Infer wavelength from a node pattern

An experimental standing-wave pattern on a string shows five antinodes between the fixed ends. The string length is

(18)

1. Which mode number is this?
2. What is the wavelength?
3. How many interior nodes are present?
4. What is the spacing between adjacent antinodes?

Exercise 12: Compare standing and traveling wave forms

Compare the equations

u1(x,t)	= 3 mm cos(4πx − 20πt),	(19)
u2(x,t)	= 6 mm cos(4πx) cos(20πt).	(20)

1. Which one is a traveling wave?
2. Which one is a standing wave?
3. Which expression has fixed nodes?
4. Which expression has a single propagation direction built into its phase?
5. If possible, identify the node positions for the standing-wave expression.

Part II: Complete Worked Solutions

Solution 1: Derive the standing-wave form

Superposition gives

u(x,t)	= u1(x,t) + u2(x,t)	(21)
	= A cos(kx − ωt) + A cos(kx + ωt).	(22)

Using

(23)

with

(24)

we obtain

(25)

1. The final expression is u(x,t) = 2A cos(kx) cos(ωt).
2. The factor depending only on position is 2A cos(kx).
3. The factor depending only on time is cos(ωt).
4. A rigidly translating wave keeps one fixed shape and shifts it through space. Here the space and time dependences are separated into a product, so the pattern oscillates in place instead of translating.

Solution 2: Read node and antinode structure from a snapshot

1. Adjacent nodes are separated by λ∕2. Therefore

   	= 0.24 m,	(26)
   λ	= 0.48 m .	(27)

2. A node is one-quarter wavelength from its nearest antinode. Thus

   (28)

3. If A₂ is at x = 0.48 m, the neighboring nodes are λ∕4 = 0.12 m away on either side:

   (29)

   Hence the neighboring nodes are at

   (30)

4. A single snapshot only shows the displacement profile at one time. A traveling wave and a standing wave can look similar in one frozen instant. To distinguish them, one must observe whether the whole profile translates or whether fixed nodes remain in place while the pattern oscillates.

Solution 3: Locate nodes and antinodes from an equation

Given

(31)

we read off

(32)

1. The coefficient is

   (33)

2. The wavelength is

   λ	=	(34)
   	= m	(35)
   	= 0.40 m .	(36)

3. For the cosine form,

   (37)

   The first three nonnegative antinodes are

   (38)

4. The positive nodes satisfy

   (39)

   The first three positive nodes are

   (40)

5. The standing wave has the form 2A cos(kx) cos(ωt) with

   (41)

   Therefore the maximum displacement magnitude at an antinode is

   (42)

Solution 4: Local amplitude at specific positions

The local amplitude magnitude is

(43)

1. At x = 0,

   (44)

   This is an antinode.

2. At x = 0.10 m,

   (45)

   This is a node.

3. At x = 0.20 m,

   (46)

   This is an antinode.

4. At x = 0.30 m,

   (47)

   This is a node.

Solution 5: Relative temporal phase of different positions

1. Points P and Q lie in the same lobe, so cos(kx) has the same sign at both. Therefore they move in the same temporal phase.
2. Point R lies in the adjacent lobe across a node, so the sign of the spatial factor reverses. Therefore P and R move in opposite temporal phase.
3. From the figure, Q is closest to an antinode, so it has the largest local amplitude.
4. Crossing a node changes the sign of cos(kx). Multiplying the same time factor by the opposite sign is equivalent to a phase shift of π in the temporal motion.

Solution 6: Recover the component traveling waves

Compare

(48)

with

(49)

Hence

(50)

1. Each component traveling wave has amplitude

   6.0 mm .

2. One possible right-moving component is

   (51)

3. One possible left-moving component is

   (52)

4. Doubling the standing-wave maximum displacement doubles 2A, so it also doubles each component-wave amplitude.

Solution 7: Determine the times of special motion

At an antinode, the position factor has magnitude 1, so the motion is simply proportional to

(53)

1. (54)

2. (55)

3. Zero displacement occurs when

   (56)

   Thus

   (57)

   so

   (58)

   In the interval 0 ≤ t ≤ 0.08 s, this gives

   (59)

4. Maximum positive displacement occurs when

   (60)

   so

   (61)

   In the interval, the times are

   (62)

5. Maximum negative displacement occurs when

   (63)

   so

   (64)

   In the interval, the times are

   (65)

Solution 8: Standing waves on a string with fixed ends

For a string fixed at both ends,

(66)

With L = 0.90 m:

1. λ1	= 2L = 1.80 m ,	(67)
   λ2	= L = 0.90 m ,	(68)
   λ3	= = 0.60 m .	(69)

2. The third mode has three half-wavelength segments, giving two interior nodes. Thus the number of interior nodes is

   (70)

3. In any standing wave, adjacent nodes are separated by λ∕2. For the second mode,

   (71)

4. Adjacent antinodes are also separated by λ∕2. For the third mode,

   (72)

Solution 9: Mode frequencies from wave speed

For fixed ends,

(73)

With v = 120 m/s and L = 0.90 m,

(74)

Therefore

f2	= 2f1 = 133.3 Hz ,	(75)
f3	= 3f1 = 200 Hz .	(76)

The ratio is

(77)

Solution 10: Diagnose conceptual statements

1. Incorrect. A standing wave does not mean every point is motionless. Only nodes remain permanently at zero displacement; points between nodes generally oscillate.
2. Incorrect. Adjacent nodes are separated by λ∕2, not λ.
3. Incorrect. Points between two neighboring nodes are in the same temporal phase, but their amplitudes vary with position and are largest at the antinode.
4. Correct. Adjacent lobes move in opposite temporal phase.
5. Incorrect. A perfect standing wave requires matched counter-propagating components with the same frequency, wavelength, and amplitude.

Solution 11: Infer wavelength from a node pattern

Five antinodes means there are five half-wavelength segments across the string, so the mode number is

(78)

Use

(79)

Thus

λ	=	(80)
	=	(81)
	= 0.50 m .	(82)

1. The mode number is

   5 .

2. The wavelength is

   0.50 m .

3. A fixed-end mode with n antinodes has n− 1 interior nodes, so the number of interior nodes is

   4 .

4. Adjacent antinodes are separated by λ∕2, so

   (83)

Solution 12: Compare standing and traveling wave forms

The two expressions are

u1(x,t)	= 3 mm cos(4πx − 20πt),	(84)
u2(x,t)	= 6 mm cos(4πx) cos(20πt).	(85)

1. u₁ is a traveling wave because its space and time appear together in a single phase combination.
2. u₂ is a standing wave because its space and time dependences are separated into a product.
3. u₂ has fixed nodes. Set

   (86)

   Hence

   (87)

   So the node positions are

   (88)

4. u₁ has a single propagation direction built into its phase, namely the right-moving form kx − ωt.
5. The node positions were found above for u₂.

Common mistakes

• Mistake: measuring a full wavelength from one node to the next. Adjacent nodes are only half a wavelength apart.
• Mistake: assuming all points in one lobe have the same displacement or amplitude. They share the same temporal phase, but not the same amplitude.
• Mistake: confusing the standing-wave maximum displacement 2A with the amplitude of each component traveling wave A.
• Mistake: assuming a single snapshot is enough to prove a pattern is standing. Time evolution matters.
• Mistake: forgetting that fixed-end string modes must satisfy the boundary condition u = 0 at both ends.

What WM10E1 reinforces

These exercises reinforce the core geometry and algebra of standing waves:

(89)

They also reinforce the ideas that nodes and antinodes are fixed in space, local amplitude varies with position, adjacent lobes oscillate in opposite temporal phase, and boundary conditions pick out the allowed normal modes. These ideas prepare the way for later treatments of resonance, harmonics, mode shapes, and energy transport.

References

References