Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 11.9, 11.54, Review Exercise 14, p. 454
A string has its tension set by a hanging mass M, and resonates at its fundamental frequency at 150 Hz. This same string will resonate with two antinodes at 150 Hz when a different mass m is hanging from it. Discuss why M > m. Explain your reasoning using the properties of wave speeds, periodic waves, and standing waves. Solution and grading rubric:
- p:
Correct. Understands how (a) the fundamental frequency is proportional to wave velocity, and (b) the wave velocity depends on the square root of the mass creating the tension, and (c) the string vibrating at its second harmonic has a fundamental frequency half that of the other string, such that the second harmonic vibrating string has a slower wave velocity, a lower tension, and thus a smaller hanging mass. - r:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. - t:
Nearly correct, but argument has conceptual errors, or is incomplete. - v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least understands how the hanging mass affects tension, and thus wave velocity. - x:
Implementation/application of ideas, but credit given for effort rather than merit. - y:
Irrelevant discussion/effectively blank. - z:
Blank.
Sections 70854, 70855
Exam code: finalPr0p
p: 18 students
r: 6 students
t: 17 students
v: 9 students
x: 0 students
y: 0 students
z: 1 student
A sample "p" response (from student 2121):
Another sample "p" response (from student 9494):
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