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About This Drill
ACT Science: Research Summaries (Drill 5) is a Science practice drill covering Research Summaries. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
ACT Research Summaries questions require you to interpret experimental data and understand how variables interact. This drill presents three experiments on standing waves in a vibrating string, varying string length, tension, and harmonic number to investigate frequency, wave speed, and wavelength.
Questions & Explanations
Experiments 1–3
Students investigated standing waves on a vibrating string. When a string fixed at both ends is driven at certain frequencies, it resonates and forms a stable pattern of nodes (points of no movement) and antinodes (points of maximum displacement). The lowest resonant frequency is called the fundamental frequency (f₁); higher resonant frequencies are called harmonics (f₂, f₃, …). Wave speed on a string is given by v = √(T/μ), where T is tension in newtons and μ is linear mass density in kg/m. Frequency and wavelength are related by v = fλ. Students used a string with μ = 0.004 kg/m (4.0 g/m) in all experiments.
Experiment 1 -- Effect of String Length on Fundamental Frequency
Tension was held constant at 4.0 N. String length was varied while the fundamental frequency was measured using a frequency meter.
Table 1. String Length vs. Fundamental Frequency (T = 4.0 N)
| Length (m) | Fundamental Frequency (Hz) |
|---|
| 0.20 | 79.1 |
| 0.25 | 63.2 |
| 0.40 | 39.5 |
| 0.50 | 31.6 |
| 0.80 | 19.8 |
| 1.00 | 15.8 |
Experiment 2 -- Effect of Tension on Wave Speed
String length was held constant at 0.50 m. Tension was varied using hanging masses, and both wave speed and fundamental frequency were recorded.
Table 2. Tension vs. Wave Speed and Fundamental Frequency (L = 0.50 m)
| Tension (N) | Wave Speed (m/s) | Fundamental Freq. (Hz) |
|---|
| 1.0 | 15.8 | 15.8 |
| 2.0 | 22.4 | 22.4 |
| 4.0 | 31.6 | 31.6 |
| 6.0 | 38.7 | 38.7 |
| 9.0 | 47.4 | 47.4 |
| 16.0 | 63.2 | 63.2 |
Experiment 3 -- Harmonics
With length fixed at 0.50 m and tension at 4.0 N (wave speed = 31.6 m/s), the string was driven at each of its first six harmonic frequencies. The corresponding wavelength of each standing wave pattern was recorded.
Table 3. Harmonic Frequencies and Wavelengths (L = 0.50 m, v = 31.6 m/s)
| Harmonic (n) | Frequency (Hz) | Wavelength (m) |
|---|
| 1 | 31.6 | 1.000 |
| 2 | 63.2 | 0.500 |
| 3 | 94.9 | 0.333 |
| 4 | 126.5 | 0.250 |
| 5 | 158.1 | 0.200 |
| 6 | 189.7 | 0.167 |
Question 1. According to Table 1, what was the fundamental frequency of the string when its length was 0.40 m?
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A) 19.8 Hz
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B) 31.6 Hz
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C) 39.5 Hz ✓
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D) 63.2 Hz
Explanation: Table 1 directly shows that at a string length of 0.40 m, the fundamental frequency was 39.5 Hz. This can also be verified using the formula f₁ = v/(2L): with wave speed v = √(4.0/0.004) = 31.6 m/s and L = 0.40 m, f₁ = 31.6/(2 × 0.40) = 31.6/0.80 = 39.5 Hz. The other values correspond to different lengths: 19.8 Hz at 0.80 m, 31.6 Hz at 0.50 m, and 63.2 Hz at 0.25 m.
Experiment 1
Students investigated standing waves on a vibrating string. When a string fixed at both ends is driven at certain frequencies, it resonates and forms a stable pattern of nodes (points of no movement) and antinodes (points of maximum displacement). The lowest resonant frequency is called the fundamental frequency (f₁); higher resonant frequencies are called harmonics (f₂, f₃, …). Wave speed on a string is given by v = √(T/μ), where T is tension in newtons and μ is linear mass density in kg/m. Frequency and wavelength are related by v = fλ. Students used a string with μ = 0.004 kg/m (4.0 g/m) in all experiments.
Experiment 1 -- Effect of String Length on Fundamental Frequency
Tension was held constant at 4.0 N.
Table 1. String Length vs. Fundamental Frequency (T = 4.0 N)
| Length (m) | Fundamental Frequency (Hz) |
|---|
| 0.20 | 79.1 |
| 0.25 | 63.2 |
| 0.40 | 39.5 |
| 0.50 | 31.6 |
| 0.80 | 19.8 |
| 1.00 | 15.8 |
Question 2. Based on Table 1, which of the following best describes the relationship between string length and fundamental frequency?
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A) As string length increases, fundamental frequency increases proportionally
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B) As string length increases, fundamental frequency decreases ✓
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C) String length has no effect on fundamental frequency
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D) As string length increases, fundamental frequency first increases then decreases
Explanation: Table 1 shows a clear inverse relationship: as string length increases from 0.20 m to 1.00 m, fundamental frequency decreases from 79.1 Hz to 15.8 Hz. This follows directly from the formula f₁ = v/(2L) -- at constant wave speed, doubling the string length halves the fundamental frequency. A longer string must accommodate one full half-wavelength across its length, so a longer string corresponds to a longer wavelength and therefore a lower frequency.
Experiment 2
Students investigated standing waves on a vibrating string. When a string fixed at both ends is driven at certain frequencies, it resonates and forms a stable pattern of nodes (points of no movement) and antinodes (points of maximum displacement). The lowest resonant frequency is called the fundamental frequency (f₁); higher resonant frequencies are called harmonics (f₂, f₃, …). Wave speed on a string is given by v = √(T/μ), where T is tension in newtons and μ is linear mass density in kg/m. Frequency and wavelength are related by v = fλ. Students used a string with μ = 0.004 kg/m (4.0 g/m) in all experiments.
Experiment 2 -- Effect of Tension on Wave Speed
String length was held constant at 0.50 m.
Table 2. Tension vs. Wave Speed and Fundamental Frequency (L = 0.50 m)
| Tension (N) | Wave Speed (m/s) | Fundamental Freq. (Hz) |
|---|
| 1.0 | 15.8 | 15.8 |
| 2.0 | 22.4 | 22.4 |
| 4.0 | 31.6 | 31.6 |
| 6.0 | 38.7 | 38.7 |
| 9.0 | 47.4 | 47.4 |
| 16.0 | 63.2 | 63.2 |
Question 3. According to Table 2, when tension increased from 1.0 N to 4.0 N, wave speed:
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A) quadrupled, from 15.8 m/s to 63.2 m/s
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B) tripled, from 15.8 m/s to 47.4 m/s
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C) doubled, from 15.8 m/s to 31.6 m/s ✓
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D) remained unchanged
Explanation: From Table 2: at 1.0 N, wave speed = 15.8 m/s; at 4.0 N, wave speed = 31.6 m/s. The speed doubled even though the tension quadrupled. This is consistent with the formula v = √(T/μ): if tension increases by a factor of 4, speed increases by √4 = 2. So doubling the speed requires quadrupling the tension -- a square-root relationship, not a linear one. This is an important distinction that the table illustrates clearly.
Experiments 1 and 2
Students investigated standing waves on a vibrating string. When a string fixed at both ends is driven at certain frequencies, it resonates and forms a stable pattern of nodes (points of no movement) and antinodes (points of maximum displacement). The lowest resonant frequency is called the fundamental frequency (f₁); higher resonant frequencies are called harmonics (f₂, f₃, …). Wave speed on a string is given by v = √(T/μ), where T is tension in newtons and μ is linear mass density in kg/m. Frequency and wavelength are related by v = fλ. Students used a string with μ = 0.004 kg/m (4.0 g/m) in all experiments.
Experiment 1 -- Effect of String Length on Fundamental Frequency
Tension was held constant at 4.0 N.
Table 1. String Length vs. Fundamental Frequency (T = 4.0 N)
| Length (m) | Fundamental Frequency (Hz) |
|---|
| 0.20 | 79.1 |
| 0.25 | 63.2 |
| 0.40 | 39.5 |
| 0.50 | 31.6 |
| 0.80 | 19.8 |
| 1.00 | 15.8 |
Question 4. Based on Table 1, if the string length were changed from 0.50 m to 0.25 m while tension remained at 4.0 N, what would happen to the fundamental frequency?
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A) It would decrease to approximately 15.8 Hz
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B) It would remain at 31.6 Hz
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C) It would increase to approximately 63.2 Hz ✓
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D) It would increase to approximately 126.4 Hz
Explanation: Table 1 already contains this data point: at L = 0.25 m, the fundamental frequency is 63.2 Hz. This can also be confirmed by reasoning: halving the string length from 0.50 m to 0.25 m doubles the fundamental frequency, since f₁ = v/(2L). At L = 0.50 m, f₁ = 31.6 Hz, so at L = 0.25 m, f₁ = 2 × 31.6 = 63.2 Hz. The answer can therefore be read directly from the table without any calculation.
Experiment 3
Students investigated standing waves on a vibrating string. When a string fixed at both ends is driven at certain frequencies, it resonates and forms a stable pattern of nodes (points of no movement) and antinodes (points of maximum displacement). The lowest resonant frequency is called the fundamental frequency (f₁); higher resonant frequencies are called harmonics (f₂, f₃, …). Wave speed on a string is given by v = √(T/μ), where T is tension in newtons and μ is linear mass density in kg/m. Frequency and wavelength are related by v = fλ. Students used a string with μ = 0.004 kg/m (4.0 g/m) in all experiments.
Experiment 3 -- Harmonics
With length fixed at 0.50 m and tension at 4.0 N (wave speed = 31.6 m/s), the string was driven at each of its first six harmonic frequencies.
Table 3. Harmonic Frequencies and Wavelengths (L = 0.50 m, v = 31.6 m/s)
| Harmonic (n) | Frequency (Hz) | Wavelength (m) |
|---|
| 1 | 31.6 | 1.000 |
| 2 | 63.2 | 0.500 |
| 3 | 94.9 | 0.333 |
| 4 | 126.5 | 0.250 |
| 5 | 158.1 | 0.200 |
| 6 | 189.7 | 0.167 |
Question 5. According to Table 3, what is the wavelength of the 4th harmonic, and how does it compare to the string length of 0.50 m?
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A) 0.500 m; equal to the string length
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B) 0.333 m; two-thirds of the string length
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C) 0.250 m; one-half of the string length ✓
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D) 0.200 m; two-fifths of the string length
Explanation: Table 3 shows the 4th harmonic has a wavelength of 0.250 m. Comparing to the string length: 0.250/0.50 = 0.5, so the wavelength is one-half the string length. This makes physical sense: the nth harmonic fits n half-wavelengths along the string, so λₙ = 2L/n. For n = 4: λ₄ = 2(0.50)/4 = 0.250 m. Each higher harmonic squeezes one more half-wavelength into the fixed string length, progressively shortening the wavelength. The pattern in Table 3 (1.000, 0.500, 0.333, 0.250, 0.200, 0.167) reflects this: each wavelength equals 1.00/n meters.