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Physics quiz question: decreased tension standing wave

Physics 205A Quiz 6, Fall Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 11.45

[Version 1]
A 0.75 m long string has a tension of 60 N. This length of string has a mass of 1.1e-4 kg. If the tension of the string were decreased by 10%, the fundamental frequency of the string would decrease by what percentage?
(A) 0%.
(B) 3.2%.
(C) 4.9%.
(D) 5.1%.

Correct answer: (D)

The fundamental frequency f_1 is given by:

f_1 = sqrt(F/mu)/(2*L) = sqrt(F*L/m)/(2*L),

where the linear mass density mu is related to the string mass and length by mu = m/L, and F is the tension in the string. If the tension in the string were decreased by 10%, then F_decreased = 0.90*F, and:

f_1,decreased = sqrt(F_decreased*L/m)/(2*L) = sqrt(0.90*F*L/m)/(2*L) = sqrt(0.90)*sqrt(F*L/m)/(2*L) = 0.949*f_1,

such that the new fundamental frequency is 1 - 0.949 = 0.051 = 5.1% less than the original fundamental frequency.

Response (A) reflects no change in the fundamental frequency. Response (B) is sqrt(10).

Student responses
Sections 70854, 70855
(A) : 2 students
(B) : 7 students
(C) : 6 students
(D) : 11 students

Student responses
Section 72177
(A) : 1 student
(B) : 3 students
(C) : 2 students
(D) : 0 students

[Version 2]
A 0.75 m long string has a tension of 60 N. This length of string has a mass of 1.1e-4 kg. If the tension of the string were increased by 10%, the fundamental frequency of the string would increase by what percentage?
(A) 0%.
(B) 3.2%.
(C) 4.9%.
(D) 5.1%.

The fundamental frequency f_1 is given by:

f_1 = sqrt(F/mu)/(2*L) = sqrt(F*L/m)/(2*L),

where the linear mass density mu is related to the string mass and length by mu = m/L, and F is the tension in the string. If the tension in the string were increased by 10%, then F_increased = 1.10*F, and:

f_1,decreased = sqrt(F_decreased*L/m)/(2*L) = sqrt(1.10*F*L/m)/(2*L) = sqrt(1.10)*sqrt(F*L/m)/(2*L) = 1.049*f_1,

such that the new fundamental frequency is 1.049 - 1 = 0.049 = 4.9% more than the original fundamental frequency.

Student responses
Sections 70854, 70855
(A) : 3 students
(B) : 11 students
(C) : 7 students
(D) : 0 students

Student responses
Section 72177
(A) : 0 students
(B) : 3 students
(C) : 3 students
(D) : 0 students

Sections 70854, 70855
Success level: 36%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.35

Section 72177
Success level: 25%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.67

2 comments:

MV said...
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MV said...

If you copy and paste to your blog after preparing a word document, you will be able to use normal symbols. This will greatly improve the use of your blog.
For instance, your post, ‘Physics quiz question: decreased tension, standing wave’, the answer can be simolified as given below:
The tension T becomes 0.9T on reducing it by 10%. So √(T) becomes √(0.9T) which is nearly 0.949√T.
Since the frequency F is given by F = (1/2L) √(T/μ), F decreases to 0.949F. The decrement is 0.051F and the percentage decrement is 5.1%.
With best wishes for your commendable service.
-MV