Cord Oscillates, Tension Changes

In summary: Keep in mind that the problem asked for the number of "complete" oscillations. So, the final answer should be 407 only if you interpret it as 407.0.In summary, the problem involves a chord stretched between two holders and gradually increasing tension. The chord oscillates in a standing wave with the fundamental frequency, and the goal is to find the number of complete oscillations within a certain time frame. By using the equation for frequency and integrating to find the total number of oscillations, the final answer is 407 complete oscillations.
  • #1
Const@ntine
285
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!
 
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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
  • #6
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?
 
  • #7
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!
 
  • #8
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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Related to Cord Oscillates, Tension Changes

1. How does the cord oscillate?

The cord oscillates due to the transfer of energy from one end of the cord to the other. When one end of the cord is pulled, it creates a wave that travels through the cord, causing it to oscillate.

2. What factors affect the oscillation of the cord?

The tension of the cord, the length of the cord, and the frequency of the pulling force all affect the oscillation of the cord. A higher tension, shorter length, and higher frequency will result in a faster and more pronounced oscillation.

3. How does tension change affect the oscillation of the cord?

An increase in tension will increase the speed and amplitude of the oscillation, as the energy transfer will be more efficient. A decrease in tension will result in a slower and smaller oscillation.

4. What happens to the cord when the tension changes?

When the tension changes, the cord will either speed up or slow down its oscillation, depending on whether the tension increases or decreases. It may also change the shape of the oscillation, such as creating a larger or smaller wave.

5. How does the cord's material affect its oscillation?

The material of the cord can affect its elasticity and density, which can impact the speed and amplitude of the oscillation. Materials with higher elasticity will result in a faster oscillation, while materials with higher density will result in a slower oscillation.

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