Max.transverse tension in a wave

[AndreAo]

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$$\pi$$f 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.

 

[Redbelly98]

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.

 

[AndreAo]

(a)The bar moves up and down 120 times per second, so the frequency f = 120 Hz.
w = 2$$\pi$$f 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:
$$\partial$$y/$$\partial$$t = -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:
$$\partial$$y/$$\partial$$x = 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!

 

[AndreAo]

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?

 

[AndreAo]

Sorry, I researched a little and see that:
P = Force x Velocity
Using this formula and replacing the values I get the result

 

[Redbelly98]

(a)The bar moves up and down 120 times per second, so the frequency f = 120 Hz.
w = 2$$\pi$$f 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:
$$\partial$$y/$$\partial$$t = -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:
$$\partial$$y/$$\partial$$x = 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!

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:

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/AnalyzingWaves.htm​

Scroll about halfway down the page, to where it says "Energy and Power in a Traveling Harmonic Wave".

 

[AndreAo]

Thanks,I find the solution for that problem:
Power = Force * Velocity
Power = (-$$\tau$$*$$\partial$$y/$$\partial$$x)*$$\partial$$y/$$\partial$$t
The negative it's because the force is contrary to the displacement?