Questions on stationary waves on strings

Question 1

Describe how a stationary wave is formed on a string.

Accepted Answer

Two progressive waves of the same frequency and amplitude travel in opposite directions (for example an incident wave and its reflection) and superpose (1), producing a pattern of nodes and antinodes that does not move along the string (1).

Question 2

State the distance between two adjacent nodes on a stationary wave, in terms of the wavelength λ.

Accepted Answer

Half a wavelength, λ/2 (1).

Question 3

Distinguish between a node and an antinode.

Accepted Answer

A node is a point of zero amplitude (no displacement) (1); an antinode is a point of maximum amplitude (1).

Question 4

The frequency of the first harmonic of a stretched string is given by f₁ = (1/2L)√(T/μ). State what the symbols T and μ represent, and give the SI unit of μ.

Accepted Answer

T is the tension in the string (1). μ is the mass per unit length of the string, measured in kg m⁻¹ (1).

Question 5

State why a stationary wave on a string fixed at both ends must have a node at each end.

Accepted Answer

The ends are fixed and cannot move, so their displacement is always zero, which is the definition of a node (1).

Question 6

Points P and Q lie between the same pair of adjacent nodes on a stationary wave. Point R lies between the next pair of nodes along. State the phase difference between P and Q, and between P and R.

Accepted Answer

P and Q: zero, they oscillate in phase (1). P and R: π rad (180°), they oscillate in antiphase (1).

Question 7

A stationary sound wave is set up in a pipe that is closed at one end and open at the other. State what the displacement pattern must have at each end, and give the reason in each case.

Accepted Answer

The closed end must be a displacement node, because the air in contact with the solid wall cannot shuttle back and forth along the tube (1). The open end must be a displacement antinode, because the air there is free and moves further than anywhere else in the pipe (1).

Question 8

A pipe of length 0.20 m is closed at one end and open at the other. Calculate the wavelength of the sound in the pipe's first harmonic.

Accepted Answer

A node sits at the closed end and an antinode at the open end, and node to antinode is a quarter of a wavelength, so the pipe holds λ/4 (1). λ = 4L = 4 × 0.20 = 0.80 m (1).

Question 9

A pipe is closed at one end and open at the other, and the air column inside it is sounding its first harmonic. State the end of the pipe at which the pressure of the air varies most, and explain your answer.

Accepted Answer

The closed end (1). The closed end is a displacement node, where the air is alternately compressed against the wall and rarefied away from it while hardly moving, so the pressure varies most there; the open end is a displacement antinode, where the air is free to move and the pressure stays close to atmospheric (1).

Question 10

A string of length 0.60 m is fixed at both ends and vibrates in its fundamental mode. The wave speed on the string is 240 m s⁻¹. Calculate (a) the wavelength and (b) the fundamental frequency.

Accepted Answer

(a) Fundamental: λ = 2L = 2 × 0.60 = 1.2 m (1) (b) f = v/λ = 240/1.2 (1) f = 200 Hz (1)

Question 11

A string of length 0.60 m is fixed at both ends. The speed of waves on the string is 240 m s⁻¹. Calculate the frequencies of the second and third harmonics.

Accepted Answer

Second harmonic: λ = L = 0.60 m (1) f = 240/0.60 = 400 Hz (1) Third harmonic: f = 3v/2L = 600 Hz (1)

Question 12

A string of length 0.80 m fixed at both ends vibrates at its fundamental frequency of 150 Hz. Calculate the speed of the waves on the string.

Accepted Answer

Fundamental: λ = 2L = 1.6 m (1) v = fλ = 150 × 1.6 (1) v = 240 m s⁻¹ (1)

Question 13

A stationary wave has a wavelength of 0.50 m. Calculate the distance between adjacent nodes, and the distance between a node and the next antinode.

Accepted Answer

Node to node = λ/2 (1) = 0.25 m (1) Node to antinode = λ/4 = 0.125 m (1)

Question 14

A guitar string has a vibrating length of 0.750 m, and this length of string has a mass of 1.20 g. The string is under a tension of 90.0 N. Calculate the frequency of the first harmonic. Use f₁ = (1/2L)√(T/μ).

Accepted Answer

μ = mass/length = (1.20 × 10⁻³)/0.750 = 1.60 × 10⁻³ kg m⁻¹ (1). f₁ = (1/(2 × 0.750)) × √(90.0/(1.60 × 10⁻³)) (1) = 158 Hz (1).

Question 15

A string fixed at both ends vibrates at its first-harmonic frequency. The length of the string and the string itself are unchanged. Determine the factor by which the tension must change for the first-harmonic frequency to double.

Accepted Answer

f₁ ∝ √T when L and μ are constant (1), so doubling f₁ requires the tension to increase by a factor of 4 (1).

Question 16

A string of length 1.2 m fixed at both ends vibrates in its fourth harmonic. State the number of nodes and the number of antinodes on the string, and calculate the wavelength.

Accepted Answer

5 nodes, counting the two fixed ends (1), and 4 antinodes (1). λ = 2L/n = (2 × 1.2)/4 = 0.60 m (1).

Question 17

A tube of length 0.85 m is open at both ends. The speed of sound in air is 340 m s⁻¹. Calculate the frequency of the first harmonic of the air column, and state the frequency of the next harmonic the tube can sound.

Accepted Answer

An open end is an antinode and antinode to antinode is half a wavelength, so λ₁ = 2L = 1.70 m (1). f₁ = v/λ = 340/1.70 = 200 Hz (1). A tube open at both ends fits every whole number of half wavelengths, so the next harmonic is the second, at 400 Hz (1).

Question 18

A tube of length 0.85 m is closed at one end. The speed of sound in air is 340 m s⁻¹. Calculate the frequency of the first harmonic of the air column and the next frequency at which the tube resonates strongly, and explain why it cannot resonate at twice its first-harmonic frequency.

Accepted Answer

A node at the closed end and an antinode at the open end fit a quarter wavelength, so λ₁ = 4L = 3.4 m (1). f₁ = 340/3.4 = 100 Hz (1). The next pattern that fits adds half a wavelength, giving the third harmonic at 300 Hz (1). Twice the first-harmonic frequency would halve the wavelength and so place one whole half wavelength in the tube; half a wavelength always runs node to node or antinode to antinode, and this tube has one end of each kind, so no even harmonic can fit (1).

Question 19

A tube of length 0.60 m is closed at one end and open at the other, and the air column in it is resonating in its third harmonic. The speed of sound in air is 340 m s⁻¹. State the number of displacement nodes and the number of displacement antinodes in the air column, and calculate the frequency at which it is resonating.

Accepted Answer

Two nodes, one at the closed end and one a further half wavelength along, and two antinodes, one of them at the open end (1). The third harmonic fits three quarter wavelengths into the tube, so λ = 4L/3 = (4 × 0.60)/3 = 0.80 m (1). f = v/λ = 340/0.80 = 425 Hz (1).

Question 20

A string of length 1.2 m fixed at both ends carries waves travelling at 180 m s⁻¹. Calculate the frequencies of the first three harmonics.

Accepted Answer

Fundamental: λ = 2L = 2.4 m (1) f₁ = v/2L = 180/2.4 = 75 Hz (1) Harmonics are multiples of f₁ (1) 75 Hz, 150 Hz, 225 Hz (1)

Question 21

A string of length 1.0 m fixed at both ends vibrates at 300 Hz, and the wave speed on it is 200 m s⁻¹. Determine which harmonic this is.

Accepted Answer

λ = v/f = 200/300 (1) λ = 0.667 m (1) Number of half-wavelengths n = 2L/λ = 2 × 1.0/0.667 (1) n = 3, so it is the third harmonic (1)

Question 22

State three differences between a progressive wave and a stationary wave.

Accepted Answer

1. A progressive wave transfers energy along its path; a stationary wave transfers no net energy (1). 2. In a progressive wave every particle has the same amplitude; in a stationary wave the amplitude varies from zero at nodes to a maximum at antinodes (1). 3. In a progressive wave the phase varies continuously with position; in a stationary wave all particles between adjacent nodes oscillate in phase (1).

Question 23

A violin string has a vibrating length of 0.328 m and a mass per unit length of 0.60 g m⁻¹. The maker states that when the tension is 50.0 N the first harmonic of the string is concert pitch A, 440 Hz. Deduce whether the maker's statement is correct. Use f₁ = (1/2L)√(T/μ).

Accepted Answer

μ = 0.60 g m⁻¹ = 6.0 × 10⁻⁴ kg m⁻¹ (1). f₁ = (1/(2 × 0.328)) × √(50.0/(6.0 × 10⁻⁴)) (1) = 440 Hz (1). This matches 440 Hz, so the maker's statement is correct (1).

Question 24

The driving frequency applied to a stretched string of fixed length 1.5 m is increased slowly. Stationary waves form at 240 Hz and next at 300 Hz, with none in between. Determine the frequency of the first harmonic and the speed of the waves on the string.

Accepted Answer

Adjacent harmonics differ by f₁, since f_n = nf₁ (1), so f₁ = 300 − 240 = 60 Hz (1). For the first harmonic λ = 2L = 3.0 m (1). v = f₁λ = 60 × 3.0 = 180 m s⁻¹ (1).

Question 25

Describe an experiment to investigate how the first-harmonic frequency of a stretched string depends on its vibrating length L. Your answer should include how the stationary wave is produced, the measurements taken, and how the results would be used to show that f₁ is inversely proportional to L.

Accepted Answer

Stretch the string from a vibration generator (driven by a signal generator) over a pulley to hanging masses, so the tension stays constant throughout (1). Adjust the driving frequency until the string vibrates in a single loop, the first harmonic, with a node at each end and one antinode at the centre (1). Record the frequency from the signal generator and measure the vibrating length L with a metre rule (1). Repeat for a range of different lengths, keeping the same string and the same tension (1). Plot f₁ against 1/L (1). A straight line through the origin shows f₁ ∝ 1/L (1).

Question 26

In a resonance-tube experiment, a tuning fork of frequency 480 Hz is held just above a vertical tube whose lower end stands in water. Lowering the water lengthens the trapped air column. Resonance is heard first when the column is 0.168 m long, and again when it is 0.523 m long. Calculate the wavelength of the sound and the speed of sound in the tube.

Accepted Answer

The column holds a quarter of a wavelength at the first resonance and three quarters at the second, so the two lengths differ by half a wavelength (1). λ = 2(L₂ − L₁) = 2(0.523 − 0.168) (1) = 0.710 m (1). v = fλ = 480 × 0.710 = 341 m s⁻¹ (1).

Question 27

Describe how the speed of sound in air is measured using a resonance tube and a tuning fork of known frequency, and explain why the method takes two resonance lengths rather than relying on the first one alone.

Accepted Answer

Stand a long tube upright with its lower end in a tall cylinder of water, so that the water closes the bottom while the top stays open, and raise or lower the tube to change the length of the trapped air column (1). Sound a fork of known frequency f a centimetre or so above the open top, without letting it touch the glass, and lengthen the column slowly until the sound swells; record that length L₁, measured from the top of the tube down to the water surface (1). Continue until a second, quieter resonance is heard and record L₂: the column is now running in its third harmonic, holding three quarters of a wavelength instead of one quarter (1). The two lengths therefore differ by half a wavelength, so λ = 2(L₂ − L₁) and v = 2f(L₂ − L₁) (1). The displacement antinode forms slightly above the open end rather than exactly at it, so every measured length is short by the same small amount; subtracting one length from the other cancels that offset, which λ = 4L₁ alone would not (1).

Question 28

A pipe sounds strongly at 340 Hz, and the next frequency at which it sounds strongly is 1020 Hz, with no resonance anywhere between the two. The speed of sound in air is 340 m s⁻¹. Deduce whether the pipe is closed at one end or open at both ends, and calculate its length.

Accepted Answer

The two resonances are in the ratio 1020/340 = 3, so the pipe sounds its first and third harmonics with the second one missing, and only a pipe with a node at one end and an antinode at the other skips the even harmonics: it is closed at one end (1). The lower frequency is therefore the first harmonic, with λ₁ = v/f = 340/340 = 1.00 m (1). A closed pipe holds a quarter of a wavelength in its first harmonic, so L = λ₁/4 (1) = 0.25 m (1).

Question 29

In a resonance-tube experiment a tuning fork is sounded above a tube standing in water, and the air column is lengthened until the first resonance is heard, at length L₁. The measurement is repeated with forks of several different frequencies f, and L₁ is plotted against 1/f. The graph is a straight line of gradient 85.0 m s⁻¹ which cuts the L₁ axis at −0.0080 m. Determine the speed of sound in the air column, and explain what the negative intercept represents.

Accepted Answer

At the first resonance the column plus a small distance e above the open end holds a quarter of a wavelength, so L₁ = v/(4f) − e, which is a straight line of gradient v/4 against 1/f (1). v = 4 × 85.0 (1) = 340 m s⁻¹ (1). The intercept is −e, so the displacement antinode forms about 8 mm beyond the open end rather than exactly at it; that end correction is the amount by which every measured length falls short (1).