Answer:
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
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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