Homework Help: Ant on a Tightrope physics problem

1. Feb 8, 2005

anaksunamoon

I'm really confused where to start on this problem. I'm not sure what I need to solve it.

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.

2. Feb 9, 2005

ehild

The ant becomes momentarily weightless when the piece of rope it stands on moves downward with acceleration g. There is a transverse wave propagating along the rope, its speed is

$$v=\sqrt{\frac{T_s}{\mu}}$$.

The frequency f can be obtained from the wavelength and from the speed as

$$f = \frac{v}{\lambda}$$

Assuming a sinusoidal wave of amplitude A, the maximum acceleration of the points of the rope is

$$a_{max} = \omega^2 A = 4\pi^2 f^2 A = g$$

ehild

3. Feb 10, 2005

anaksunamoon

Little Problem

Okay since I'm solving for a_max, I need to use the variables g, f, A.

What I did was I inserted the values (v/lambda) for f in the equation 4*PI^2*f^2*A = g, then I inserted sqrt(T_s/mu) for "v". Then I divided that whole thing by g and I got it wrong.

This is what I came up with

(4*(PI^2)*A*(((sqrt(T_s/mu))*(1/lambda))^2))/g

Should it be negative since it's going downward? I think I'm on the right track, but I'm not sure. What does finding the maximum have to do with finding the minimum?

(btw, thanks echild)

4. Feb 11, 2005

ehild

You have to find the minimum amplitude which yields an acceleration of maximum magnitude equal to g. The ant is shaken off at the moment when the rope accelerates downward below all its six feet and this acceleration is at least as big as g. I hope you understand what I mean :).

Do you remember the time dependence of the displacement, velocity and acceleration of a point which performs harmonic vibration? It is something like

$$y = A \cos (\omega* t)$$,

the velocity is $$v = -A \omega \sin (\omega* t)$$,

and the acceleration is

$$a = -A \omega^2 \cos (\omega* t)$$

A is the amplitude of the wave travelling along the rope, and also the amplitude of vibration at the place of the ant. You have to find an appropriate A (how much you have to shake the rope) so as the acceleration some time during the vibration should be equal to g.

You see that the maximum acceleration during the vibration is

$$a_{max} = A* \omega^2 = 4\pi^2 f^2$$.

It is enough if only the maximum of the acceleration is as big as g: the ant is shaken off at the moment when the acceleration reaches this value. To get this value, the amplitude must exceed some minimum and you have to determine this minimum value of the amplitude.

.

What does this stand for?

You need to give an equation in the form: "the minimum amplitude should be $$A_{min} = F(\lambda, \mu, T_s, g)$$"

Anyway, do not forget that you have f^2, so the square root will cancel.

ehild

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