Alarge ant is standing on the middle of a circus tightrope that is stretched with tension ts. the rope has mass per unit length μ. 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 λ and amplitude a. assume that the magnitude of the acceleration due to gravity is g. what is the minimum wave amplitude amin 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 ts, μ, λ, and g.

The minimum wave amplitude in terms of Ts, μ, λ, and g is;

A = λ²gμ/(4π²Ts)

        Since we are told that the large ant stands on a rope that is stretched, then the position motion of the ant would be represented by a sinusoidal wave:

y = A sin(ωt)

         We are told that the ant becomes weightless when its acceleration is equal to gravity. Let us find the acceleration which is the second derivative of the position equation to get;

y' = Aω cos(ωt)

y" = -Aω² sin(ωt)

Acceleration; a = -Aω² sin(ωt)

          Now, the maximum acceleration will occur when sin(ωt) = 1 .  This will happen when a = -g.  

Thus;

-g = -Aω²

A = g/ω²

         Formula for angular frequency in terms of velocity and wavelength is;

ω = 2πv/ λ

Thus;

A = g/(2πv/λ)²

A = λ²g/(2λπ)²v²

        Now, the formula for velocity of the wave in terms of tension Ts and linear density μ is:

v = √(Ts/μ)

Thus;

v² = Ts/μ

Thus;

A = λ²g/((2π)²(Ts/μ))

A = λ²gμ/(4π²Ts)

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A = g (λ / 2π)² μ / Ts

Explanation:

The ant becomes "weightless" when its acceleration is equal to gravity.  The motion of the ant is a sinusoidal wave:

y = A sin(ωt)

By taking the derivative twice, we can find the acceleration of the ant:

y' = Aω cos(ωt)

y" = -Aω² sin(ωt)

The maximum acceleration occurs when sine is 1.  We want this to happen at a = -g.

-g = -Aω² (1)

A = g / ω²

Angular frequency is 2π times the normal frequency:

ω = 2πf

A = g / (2πf)²

Frequency is velocity divided by wavelength:

f = v / λ

A = g / (2πv / λ)²

A = g (λ / 2π)² / v²

Velocity of a wave in a string with tension Ts and linear density μ is:

v = √(Ts / μ)

Therefore:

v² = Ts / μ

Plugging in:

A = g (λ / 2π)² / (Ts / μ)

A = g (λ / 2π)² μ / Ts

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