Max.transverse tension in a wave

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A transverse sinusoidal wave is generated on a string with a linear density of 120 g/m and a tension of 90 N, producing a maximum transverse speed of 3.77 m/s, which slightly differs from the book's value of 3.66 m/s. The calculations for angular frequency and wave number were derived correctly, with the wave equation established as y(x,t)=0.005sin(27.49x - 753.98t). The maximum transverse component of tension was calculated to be 12.3 N, close to the book's answer of 12.2 N. The discussion also touched on finding the maximum power transferred along the string, leading to the formula Power = Force x Velocity, with further exploration of energy in traveling waves.
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Homework Statement


A transverse sinusoidal wave is generated at one end of a long, horizontal string by a bar that moves up and down through a distance of 1.00cm. The motion is continuous and is repeated regularly 120 times per second. The string has linear density 120 g/m and is kept under a tension of 90.0N. Find (a) the maximum value of the transverse speed u and (b) the maximum value of the transverse component of the tension.

Homework Equations


y(x,t)=A sin(kx-wt) traveling wave
k = \frac{2\pi}{\lambda} angular wave number
w = 2\pif angular frequency
v = \sqrt{\frac{T}{\mu}} wave speed
\mu linear density

The Attempt at a Solution


After calculating the values of k and w, and deriving the first equation, I found u = 3.77 m/s the answer in the book say it is 3.66 m/s is anything wrong?
Letter b I have no idea on how to start solving. I tried to calculate it using F = ma, but I don't know how much mass I should consider, and probably this way is not right.
 
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AndreAo said:

The Attempt at a Solution


After calculating the values of k and w, and deriving the first equation, I found u = 3.77 m/s the answer in the book say it is 3.66 m/s is anything wrong?
Letter b I have no idea on how to start solving. I tried to calculate it using F = ma, but I don't know how much mass I should consider, and probably this way is not right.

(a) Can you show your calculations of k and w? That would help us figure out what is wrong.

(b) First you'll need to figure out what is the maximum angle the string makes with the horizontal direction.
 
(a)The bar moves up and down 120 times per second, so the frequency f = 120 Hz.
w = 2\pif replacing f = 120 Hz, w = 2*\pi*120 = 753,98 s^{-1}. How k = w/v and v = \sqrt{\frac{T}{\mu}} replacing the value of T = 90 N and \mu = 0,12 kg/m I received v = 27,39 m/s.
k = 752,98/27,39
k = 27,49 m^{-1}
With these values(and amplitude = 0.05) the equation is:
y(x,t)=0,005sin(27,49x - 753,98t)
Deriving in t:
\partialy/\partialt = -0,005*753,98cos(27,49x - 753,98t) so the max. value of velocity happens when cos -1 and u = 3,77 m/s.

(b)The tan of this max.angle:
\partialy/\partialx = 0,138 cos(27,49x - 753,98t), which max.value is 0,138. arctan 0,138 = 7,85 degrees, so the angle between the tension and the vertical is 90 - 7,85 = 82,15 degrees. Using T_{v}= cos \theta * T
T_{v} = cos 82,15 * 90
T_{v} =12,3 N, book's answer is 12,2N, i think it's right :D
Thanks a lot!
 
Yet in this exercise, it asks me what's the maximum power transferred along the string. I know the formula of kinetic energy, but I can't figure out what is the formula for potencial energy on this string. Do I have to first find this formula before calculating the max.value of power? How can I do this?
 
Sorry, I researched a little and see that:
P = Force x Velocity
Using this formula and replacing the values I get the result
 
AndreAo said:
(a)The bar moves up and down 120 times per second, so the frequency f = 120 Hz.
w = 2\pif replacing f = 120 Hz, w = 2*\pi*120 = 753,98 s^{-1}. How k = w/v and v = \sqrt{\frac{T}{\mu}} replacing the value of T = 90 N and \mu = 0,12 kg/m I received v = 27,39 m/s.
k = 752,98/27,39
Remember that k=753,98. This will not make a big difference though.
k = 27,49 m^{-1}
With these values(and amplitude = 0.05) the equation is:
y(x,t)=0,005sin(27,49x - 753,98t)
Deriving in t:
\partialy/\partialt = -0,005*753,98cos(27,49x - 753,98t) so the max. value of velocity happens when cos -1 and u = 3,77 m/s.
Looks good to me, I don't know why your book got a different number. Note also that the answer is (amplitude)*(w), so k did not really enter into this calculation.

(b)The tan of this max.angle:
\partialy/\partialx = 0,138 cos(27,49x - 753,98t), which max.value is 0,138. arctan 0,138 = 7,85 degrees, so the angle between the tension and the vertical is 90 - 7,85 = 82,15 degrees. Using T_{v}= cos \theta * T
T_{v} = cos 82,15 * 90
T_{v} =12,3 N, book's answer is 12,2N, i think it's right :D
Thanks a lot!
Looks good, you're welcome!

AndreAo said:
Yet in this exercise, it asks me what's the maximum power transferred along the string. I know the formula of kinetic energy, but I can't figure out what is the formula for potencial energy on this string. Do I have to first find this formula before calculating the max.value of power? How can I do this?
Try this page:


Scroll about halfway down the page, to where it says "Energy and Power in a Traveling Harmonic Wave".
 
Thanks,I find the solution for that problem:
Power = Force * Velocity
Power = (-\tau*\partialy/\partialx)*\partialy/\partialt
The negative it's because the force is contrary to the displacement?
 

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