String Harmonics Quiz: Explore Vibrating Strings Concepts

1. For a fixed–fixed string, the (n)-th harmonic frequency is approximately (f_n = n f_1).

True

False

Concept: harmonic series on an ideal string. Each higher mode fits an integer number of half-wavelengths. That produces frequencies that are integer multiples of the fundamental.

Explanation

Concept: harmonic series on an ideal string. Each higher mode fits an integer number of half-wavelengths. That produces frequencies that are integer multiples of the fundamental.

2. The best overall summary is:

Harmonics on strings are random

For a fixed–fixed string, λ_n=2l/n) and (f_n=nf_1); changing length or tension changes pitch

Plucking harder changes wave speed

Higher harmonics have lower frequency

Concept: string harmonics recap. Integer-mode patterns give predictable wavelengths and frequencies. Pitch control comes from changing length, tension, or mass per length.

Explanation

Concept: string harmonics recap. Integer-mode patterns give predictable wavelengths and frequencies. Pitch control comes from changing length, tension, or mass per length.

3. Harmonics occur because only certain standing-wave patterns satisfy the boundary conditions.

True

False

Concept: allowed modes. Fixed ends require nodes at specific locations. Only waves that fit those constraints persist as resonant modes.

Explanation

Concept: allowed modes. Fixed ends require nodes at specific locations. Only waves that fit those constraints persist as resonant modes.

4. If the fundamental wavelength is (2l), then the fundamental frequency is (f_1 = v/2l), where (v) is wave ______.

Concept: using (v=f λ). Frequency depends on wave speed and allowed wavelength. String properties set (v), while boundary conditions set λ.

Explanation

Concept: using (v=f λ). Frequency depends on wave speed and allowed wavelength. String properties set (v), while boundary conditions set λ.

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5. In the second harmonic, there is a node at the center of the string.

True

False

Concept: node pattern. The second harmonic has nodes at both ends and one additional node in the middle. This divides the string into two equal vibrating sections.

Explanation

Concept: node pattern. The second harmonic has nodes at both ends and one additional node in the middle. This divides the string into two equal vibrating sections.

6. The third harmonic on a fixed–fixed string has how many “loops” (antinodes)?

1

2

3

4

Concept: mode shape counting. The (n)-th harmonic has (n) antinodes (loops) on a fixed–fixed string. The third harmonic has three loops.

Explanation

Concept: mode shape counting. The (n)-th harmonic has (n) antinodes (loops) on a fixed–fixed string. The third harmonic has three loops.

7. Plucking harder mainly increases amplitude (loudness), not the harmonic frequencies.

True

False

Concept: amplitude vs frequency. In linear systems, frequency is set by the system, not how hard you pluck. Harder plucks add energy, increasing amplitude and often altering harmonic mix slightly, but not the ideal frequencies.

Explanation

Concept: amplitude vs frequency. In linear systems, frequency is set by the system, not how hard you pluck. Harder plucks add energy, increasing amplitude and often altering harmonic mix slightly, but not the ideal frequencies.

8. Which change would lower a string’s fundamental frequency?

Shortening the string

Increasing tension

Increasing the vibrating length

Using the same string but plucking harder

Concept: what sets fundamental frequency. (f_1) decreases when (l) increases or when wave speed decreases. Plucking harder changes amplitude, not the natural frequencies (in linear behavior).

Explanation

Concept: what sets fundamental frequency. (f_1) decreases when (l) increases or when wave speed decreases. Plucking harder changes amplitude, not the natural frequencies (in linear behavior).

9. If you increase tension in a string, the wave speed increases and the fundamental frequency increases.

True

False

Concept: tension raises pitch. Higher wave speed means higher frequencies for the same wavelength conditions. That is why tightening a string raises pitch.

Explanation

Concept: tension raises pitch. Higher wave speed means higher frequencies for the same wavelength conditions. That is why tightening a string raises pitch.

10. If a string’s fundamental is 200 Hz, the fifth harmonic is:

400 Hz

600 Hz

800 Hz

1000 Hz

Concept: harmonic calculation. Fifth harmonic is (5f_1). (5 *200 = 1000) Hz.

Explanation

Concept: harmonic calculation. Fifth harmonic is (5f_1). (5 *200 = 1000) Hz.

11. If (v) is the wave speed, the fundamental frequency of a fixed–fixed string is:

(f_1 = v/l)

(f_1 = v/(2l))

(f_1 = 2v/l)

(f_1 = lv)

Concept: fundamental frequency formula. From λ_1=2l and (v=f λ), (f_1=v/(2l)). This is a key relation in string instruments.

Explanation

Concept: fundamental frequency formula. From λ_1=2l and (v=f λ), (f_1=v/(2l)). This is a key relation in string instruments.

12. Fretting a guitar string (shortening its vibrating length) raises the pitch mainly by increasing the fundamental frequency.

True

False

Concept: guitar pitch control. When you press the string against a fret, you effectively reduce (l). That increases (f_1), producing a higher note.

Explanation

Concept: guitar pitch control. When you press the string against a fret, you effectively reduce (l). That increases (f_1), producing a higher note.

13. The wave speed on a string depends on the string’s tension and mass per length.

True

False

Concept: wave speed on strings. Higher tension generally increases wave speed. Higher mass per length reduces wave speed.

Explanation

Concept: wave speed on strings. Higher tension generally increases wave speed. Higher mass per length reduces wave speed.

14. If the fundamental frequency is 110 Hz, the third harmonic is:

55 Hz

220 Hz

330 Hz

440 Hz

Concept: integer multiples. The third harmonic is (3f_1). So (3 \times 110 = 330) Hz.

Explanation

Concept: integer multiples. The third harmonic is (3f_1). So (3 \times 110 = 330) Hz.

15. For a string fixed at both ends, the fundamental wavelength is:

(l)

(2l)

(l/2)

(4l)

Concept: fundamental wavelength on a string. The fundamental mode fits half a wavelength into the length: (l=\lambda/2). So λ_1=2l.

Explanation

Concept: fundamental wavelength on a string. The fundamental mode fits half a wavelength into the length: (l=\lambda/2). So λ_1=2l.

16. For the (n)-th harmonic on a fixed–fixed string, the wavelength is λ_n = 2l/{______}.

Concept: wavelength pattern. The string length contains (n) half-wavelengths: (l=nλ/2). Rearranging gives λ_n=2l/n.

Explanation

Concept: wavelength pattern. The string length contains (n) half-wavelengths: (l=nλ/2). Rearranging gives λ_n=2l/n.

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17. The spacing between harmonics in frequency for an ideal string is constant (equal to (f_1)).

True

False

Concept: harmonic spacing. Since (f_n = n f_1), consecutive harmonics differ by (f_1). This produces evenly spaced frequencies.

Explanation

Concept: harmonic spacing. Since (f_n = n f_1), consecutive harmonics differ by (f_1). This produces evenly spaced frequencies.

18. If wave speed on a string stays constant and you shorten the vibrating length, the fundamental frequency:

Decreases

Increases

Stays the same

Becomes zero

Concept: frequency vs length. For the fundamental, (f_1 = v/(2l)). Smaller (l) makes (f_1) larger, giving a higher pitch.

Explanation

Concept: frequency vs length. For the fundamental, (f_1 = v/(2l)). Smaller (l) makes (f_1) larger, giving a higher pitch.

19. For a fixed–fixed string, the second harmonic wavelength is λ_2 = ______.

Concept: second harmonic wavelength. The second harmonic fits one full wavelength into the length: (l= λ _2). So λ_2=l.

Explanation

Concept: second harmonic wavelength. The second harmonic fits one full wavelength into the length: (l= λ _2). So λ_2=l.

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20. If you halve the string length (same tension and string), the fundamental frequency becomes:

Half as large

Twice as large

Four times as large

Unchanged

Concept: inverse relationship with length. Since (f_1=v/(2l)), halving (l) doubles (f_1). This is why shorter strings sound higher.

Explanation

Concept: inverse relationship with length. Since (f_1=v/(2l)), halving (l) doubles (f_1). This is why shorter strings sound higher.

String Harmonics Quiz: Test Your Understanding of Vibrating Strings

1. For a fixed–fixed string, the (n)-th harmonic frequency is approximately (f_n = n f_1).

2.

What first name or nickname would you like us to use?

2. The best overall summary is:

3. Harmonics occur because only certain standing-wave patterns satisfy the boundary conditions.

4. If the fundamental wavelength is (2l), then the fundamental frequency is (f_1 = v/2l), where (v) is wave ______.

5. In the second harmonic, there is a node at the center of the string.

6. The third harmonic on a fixed–fixed string has how many “loops” (antinodes)?

7. Plucking harder mainly increases amplitude (loudness), not the harmonic frequencies.

8. Which change would lower a string’s fundamental frequency?

9. If you increase tension in a string, the wave speed increases and the fundamental frequency increases.

10. If a string’s fundamental is 200 Hz, the fifth harmonic is:

11. If (v) is the wave speed, the fundamental frequency of a fixed–fixed string is:

12. Fretting a guitar string (shortening its vibrating length) raises the pitch mainly by increasing the fundamental frequency.

13. The wave speed on a string depends on the string’s tension and mass per length.

14. If the fundamental frequency is 110 Hz, the third harmonic is:

15. For a string fixed at both ends, the fundamental wavelength is:

16. For the (n)-th harmonic on a fixed–fixed string, the wavelength is λ_n = 2l/{______}.

17. The spacing between harmonics in frequency for an ideal string is constant (equal to (f_1)).

18. If wave speed on a string stays constant and you shorten the vibrating length, the fundamental frequency:

19. For a fixed–fixed string, the second harmonic wavelength is λ_2 = ______.

20. If you halve the string length (same tension and string), the fundamental frequency becomes: