Standing Wave on a Driven String
A string of length 0.650 m between two fixed supports has a linear density of 3.2 g/m and is under 80 N of tension. A student drives the string with a mechanical oscillator and searches only from 300 Hz to 450 Hz. At 365 Hz the string shows a large-amplitude standing wave with three antinodes and four nodes, including the endpoints. The student calls this the fundamental frequency because it is the lowest frequency that produced a clear pattern in the search range. The oscillator’s frequency is accurate to ±1 Hz, and the tension is set with a spring scale accurate to ±2 N. The string’s maximum safe tension is 250 N. A second student says the pattern is a higher harmonic and that the fundamental was missed because the search started too high.
1. Calculate the fundamental frequency and identify the harmonic number of the 365 Hz pattern. Compare your result with the student’s claim.
Read the worked answer
Convert the linear density:
μ = 3.2 g/m = 3.2 × 10⁻³ kg/m.
The wave speed is
v = sqrt(T/μ) = sqrt(80 / 3.2 × 10⁻³) = sqrt(2.50 × 10⁴) = 158 m/s.
For a string fixed at both ends,
f_n = n v / (2L).
The fundamental is
f_1 = 158 / (2 × 0.650) = 122 Hz.
The observed frequency is
365 / 122 = 2.99 ≈ 3,
so 365 Hz is the third harmonic, or f_3 ≈ 366 Hz. The three-antinode pattern is exactly the n = 3 standing wave. The tension uncertainty of ±2 N changes the speed by about 1.3%, so the predicted f_3 is still about 365–366 Hz.
The second student is correct. The fundamental near 122 Hz was outside the search range, so the student did not find the lowest possible standing-wave frequency. The 365 Hz pattern is not the fundamental.
2. Determine the tension needed to make the fundamental frequency 240 Hz. Is that tension safe for this string? What alternatives exist?
Read the worked answer
Starting from
f_1 = (1/(2L)) sqrt(T/μ),
solve for tension:
T = μ(2L f_1)².
With L = 0.650 m and f_1 = 240 Hz,
2L f_1 = 2(0.650)(240) = 312 m/s.
Then
T = (3.2 × 10⁻³)(312)² = (3.2 × 10⁻³)(97,344) = 311.5 N.
This is about 312 N, which exceeds the string’s 250 N safe limit by about 25%. Therefore the string cannot be tuned to a 240 Hz fundamental at this length and linear density without exceeding the safe tension.
Alternatives include using a shorter vibrating length, using a string with lower linear density, or accepting a different harmonic. The observation at 365 Hz already corresponds to the third harmonic; forcing the fundamental to 240 Hz would require rebuilding the string’s length or mass density, not merely adjusting tension within the safe range.
3. Suppose the student had counted two antinodes instead of three at 365 Hz. How would that change the interpretation, and what fundamental would then be implied?
Read the worked answer
If the pattern at 365 Hz had exactly two antinodes, it would be the second harmonic, so
f_2 = 365 Hz
and the implied fundamental would be
f_1 = 365/2 = 182.5 Hz.
That would require the wave speed to be
v = 2L f_1 = 2(0.650)(182.5) = 237 m/s.
With the given linear density, the required tension would be
T = μv² = (3.2 × 10⁻³)(237)² = 180 N,
which is within the 250 N safe limit. This is very different from the three-antinode interpretation, which gives f_1 ≈ 122 Hz and requires about 312 N for a 240 Hz fundamental. The lesson is that the node and antinode count must be identified carefully, because a miscount changes the harmonic number and therefore the inferred fundamental and string properties.