Find min. wave amplitude (A_min) for Ant on Tightrope: T_s, mu, lambda, g

[ussrasu]

A large ant is standing on the middle of a circus tightrope that is stretched with tension (T_s). The rope has mass per unit length (mu). Wanting to shake the ant off the rope, a tightrope walker moves her foot up and down near the end of the tightrope, generating a sinusoidal transverse wave of wavelength (lambda) and amplitude (A) . Assume that the magnitude of the acceleration due to gravity is (g).

What is the minimum wave amplitude (A_min) such that the ant will become momentarily weightless at some point as the wave passes underneath it? Assume that the mass of the ant is too small to have any effect on the wave propagation.
Express the minimum wave amplitude in terms of T_s, mu, lambda, and g.

How would i go about solving this question? i don't know where to begin? Any help would be much appreciated!

 

[quasar987]

By "momentarily weightless", your teacher probably meant "the net acceleration will be momentarily 0". You know the shape of the traveling wave is sinusoidal. Can you write its general equation y(x,t)=...?

If you can, then you can differentiate that 2 times wrt time and that'll give you the vertical acceleration of the rope as a function of position and time. Set that acceleration equal to 'g', then solve for A and "minimize" it.

 

[Dr.Brain]

Ok This has to be thought upon.

First ask yourself when will the ant feel weightless?...when going up with acceleration or going down with some acceleration.Solution lies in going down with acceleration because this acceleration of going down will oppose the g.

Therefore the ant should go down with acceleration "g" so that it ffels weightless"

Now suppose the sinusoidal wave created by the man by tapping his foot is given by:

y=Asinwt

Now double differentiate it wrt t so that you get the acceleration at which the rope particles go up and down:

a=-Aw^2 sinwt

Now we know a=g , for ant to feel weight less.

Therefore,

g=Aw^2sinwt

Now for A to be minimum ,sinwt=1

Therefore you get

g=Aw^2

Now for the value of "w" , we need the wavelength.

Now velocity of the sinusoidal wave is given by:

v= \sqrt \frac{T}{m}

where m= mass per unit length

Now apply v=f (lamda)

Calculate frequency from above formula

Now w= 2(pie)f

Put the value in above equation to get the answer...easy isn't it?

 

[ussrasu]

Yes, thanks a lot for the help Dr Brain

 

[quasar987]

What I find strange is that the velocity of the wave is unafected by the fact that the force of gravity acts on the rope. The equation of motion for an "element" of rope is affected, therefor the resulting wave equation should be affected... Unless we neglect the force of gravity because it's so small compared to the tension... which is probably what we do.

 

[hotmail590]

g/ (  (  (   (2pi)(sqrt(T_s/mu)   )   /   (lambda)    )^2 )

that is what i got for A min, would that be correct?