String waves on a cable car and free falling.

  • Thread starter Gravitino22
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  • #1
Gravitino22
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Homework Statement



You're riding a cable car from Bogota up to Monser-
rate. During your ride there is a small earthquake that
sends vertical, transverse waves propagating along the
cable. The cable's tension is T, its linear mass den-
sity is u, and the wavelength of the wave is [tex]\lambda[/tex]. If the
amplitude of the wave is large enough, the motion of
the cable car will momentarily leave you in free fall.
Show that in order for this to happen the amplitude
must be given by

Y => [tex]\frac{g\lambda^{2}u}{4\pi^{2}T}[/tex]

Homework Equations


position along a wave: y=Ysin(kx-wt)

string waves speed: v = [tex]\sqrt{\frac{T}{u}}[/tex]



The Attempt at a Solution



My train of thought is that for free fall to happen the acceleration of the wave must be greater than the acceleration of gravity.

I tried taking the second partial derivate with respect to time of the waves position equation to obtain its acceleration which gave me:

a[tex]_{y}[/tex]=-w[tex]^{2}[/tex]Ysin(kx-wt) that should be => then g

but i don't know where the tension would or the linear mass density would come in.


Any help pointing me in the right direction would be appriciated! thanks :)
 

Answers and Replies

  • #2
willem2
2,112
377
a[tex]_{y}[/tex]=-w[tex]^{2}[/tex]Ysin(kx-wt) that should be => then g

but i don't know where the tension would or the linear mass density would come in.

try to compute k and w.
 
  • #3
Gravitino22
30
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w = 2[tex]\pi[/tex]f where f is the angular frequency and k =[tex]\frac{2\pi}{\lambda}[/tex]

that would give you sin(2[tex]\pi[/tex]([tex]\frac{x}{\lambda}[/tex] - ft)) by factoring a 2pii inside... still not seeing how the sine term is somehow related to the tension and linear mass density =/
 
  • #4
willem2
2,112
377
the frequency depends on the wave speed and the wavelength
 
  • #5
Gravitino22
30
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ok i made the following substitutions:

f=[tex]\frac{v}{\lambda}[/tex] and made v=[tex]\frac{x}{t}[/tex] (iam unsure about this one)


so far i have this

g<=-[tex]\frac{4\pi^{2}x^{2}Y}{\lambda^{2}t^{2}}[/tex]sin(2[tex]\pi[/tex]) the sin term is 1.
 
  • #6
Gravitino22
30
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wait wait! i have it just substitute back for v[tex]^{2}[/tex] and which equals T/u and viola!


tho there's a small problem...the negative sign...


Thanks a lot tho man!
 
  • #7
Gravitino22
30
0
nm! i don't have it now...iam stupid sine of 2pii is not 1 lol that sine term still bothering me...somehow i know i have to make the stuff inside the sine equal to pii/2.
 
  • #8
willem2
2,112
377
nm! i don't have it now...iam stupid sine of 2pii is not 1 lol that sine term still bothering me...somehow i know i have to make the stuff inside the sine equal to pii/2.

you already gave an equation for v.

For the amplitude you can just take the maximum value of sin() which is 1
 
  • #9
Gravitino22
30
0
hmm yea that's what i was thinking...Thanks a lot :)
 

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