Cord Oscillates, Tension Changes

Click For Summary

Homework Help Overview

The problem involves a chord with a specified linear density that is stretched between two holders, with a tension that changes over time. The task is to determine the number of complete oscillations the chord undergoes as the tension varies from 15.0 N to 25.0 N over a period of 3.50 seconds, while oscillating in a fundamental mode.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between tension and frequency, exploring the implications of a changing tension on the oscillation frequency. There are attempts to derive expressions for the number of oscillations based on the fundamental frequency and the changing tension. Some participants question the integration process needed to find the total number of oscillations over the given time period.

Discussion Status

The discussion has progressed with participants providing insights into the mathematical relationships involved. Some have suggested integration as a necessary step to find the total number of oscillations, while others have worked through specific calculations and expressed uncertainty about their results. There is acknowledgment of a potential resolution as one participant arrives at a number that matches a reference answer, though questions about the process remain.

Contextual Notes

Participants note the complexity introduced by the changing tension and the need to apply calculus to solve the problem. There is an emphasis on ensuring proper integration techniques and handling of variables throughout the discussion.

Const@ntine
Messages
285
Reaction score
18

Homework Statement



A chord with linear density μ = 0.00160 kg/m, is stretched between two holders, which have a distance of 0.480 m between them (so the length of the chord is L = 0.480 m). The chord doesn't stretch enough to notice, when the tension T gradually goes from 15.0 N at t = 0s, to 25.0 N, at t = 3.50 s. So, T = 15.0 N + 10.0 kgm/s3*t/3.50. During that time, the chordoscillates with the fundemental, normal way of oscillation. How many complete oscillations will it cover in that time?

Homework Equations



v = λ*f
v = sqrt(T/μ)

The Attempt at a Solution



Dunno what to do here, really. When I heard fundemental, I figured it was a case of standing waves, so I went ahead and tried finding the frequency (from the formula f = n/2L * sqrt(T/μ), with n = 1) at 0s, then at 3.50s, finding the average f, then finding the Period T (f = 1/T), and finding how many times T fits in the 3.50s timespan. But obviously that was wrong.

To be fair, I don't know what to do here. I've never seen anything like this, with the Tension gradually chaning, and the chord being described as merely oscillating instead of producing a standing wave or something.

Any help is appreciated!
 
Physics news on Phys.org
The chord is oscillating in a standing wave.

Think in terms of infinitesimals. If dN is the number of oscillations during a time dt, how would you express dN in terms of f and dt?
 
TSny said:
The chord is oscillating in a standing wave.

Think in terms of infinitesimals. If dN is the number of oscillations during a time dt, how would you express dN in terms of f and dt?

Technically the definition of frequency is the number of revisions/oscillations divided by the timespan, f = N/Δt. For an infinitely small timeframe, logically, f = dN/dt.

Now, it oscillates as a standing wave, with the fundamental frequency, so f = 1/2L * sqrt(T/μ), right?
 
Darthkostis said:
Technically the definition of frequency is the number of revisions/oscillations divided by the timespan, f = N/Δt. For an infinitely small timeframe, logically, f = dN/dt.

Now, it oscillates as a standing wave, with the fundamental frequency, so f = 1/2L * sqrt(T/μ), right?
Yes. You know T as a function of t.
 
TSny said:
Yes. You know T as a function of t.
So I have:

dN/dt = 1/2L * sqrt(15.0 + 10.0dt/3.50 / 0.00160)

How does that help me though? Will I have to square both sides of the equation?

(dN/dt)2 = 1/4L2 * ((15 + 10.0dt/3.5)/0.00160)

If I put all the numbers in (L = 0.480), I get dN = sqrt(10168.1 + 1938.7dt)*dt

If I put in dt = t = 3.50s, then N = 457, which is different from the book's answer, 407. Did I make an error in the multiplications and whatnot, or was my whole line of thinking wrong?
 
Rearrange dN/dt = f to get dN = f dt. In this problem f is a function of time.

Have you studied calculus? If so, what calculus operation would you apply to dN = f dt in order to get the total number of oscillations, N, that occur between t = 0 and t = 3.5 s?
 
TSny said:
Rearrange dN/dt = f to get dN = f dt. In this problem f is a function of time.

Have you studied calculus? If so, what calculus operation would you apply to dN = f dt in order to get the total number of oscillations, N, that occur between t = 0 and t = 3.5 s?

Oh yeah, I just have to integrate it, darn it. I thought about it when I saw the formula for T, but I figured it was too complex with the square root and all that. Anyway:

dN = ∫3.50 (1/2L * sqrt(T/μ))*dt = ... = 1/0.960 ∫3.50 sqrt(9375 + 1787.5t)*dt

I set 9375 + 1787.5t = u => dt = du/1787.5

dN = 1/1716 ∫3.50 u1/2 du = 1/1716[2/3 * u3/2]3.50

I didn't modify the edges of the integral, so where I have u, I replace it with (9375 + 1787.5t) and do the math.

In the end I get 406.6, which rounded up is 407, the book's answer. Any mistakes are just due to the Significant Digits during the math.

Thanks a ton for the help, I really appreciate it!
 
Darthkostis said:
dN = 1/1716 ∫3.50 u1/2 du = 1/1716[2/3 * u3/2]3.50

I didn't modify the edges of the integral, so where I have u, I replace it with (9375 + 1787.5t) and do the math.
You integrate both sides of the equation, so the left side is not dN, but ∫dN.

In the end I get 406.6, which rounded up is 407, the book's answer. Any mistakes are just due to the Significant Digits during the math.

Thanks a ton for the help, I really appreciate it!
OK. Good work.
 
  • Like
Likes   Reactions: Const@ntine

Similar threads

Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 70 ·
3
Replies
70
Views
6K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K