Cord Oscillates, Tension Changes

AI Thread Summary
The discussion revolves around calculating the number of complete oscillations of a chord with a gradually changing tension from 15.0 N to 25.0 N over 3.50 seconds. The fundamental frequency is derived from the formula f = (1/2L) * sqrt(T/μ), where T is a function of time. The user initially struggles with the concept of integrating to find the total number of oscillations but eventually realizes that integrating the frequency function over the time interval is necessary. After performing the integration, the user arrives at a total of approximately 407 oscillations, which matches the book's answer. The conversation highlights the importance of calculus in solving physics problems involving variable tension.
Const@ntine
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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!
 
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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.
 
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