Question

    A large ant is standing on the middle of acircus tightrope that is stretched with tension T_s. The rope has mass per unit length μ. Wanting to shake the ant off the rope, a tightropewalker 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 waveamplitude A_min such that the ant will become momentarily weightless atsome point as the wave passes underneath it? Assume that the massof the ant is too small to have any effect on the wavepropagation.

 

Express the minimum wave amplitude interms of T_s, μ, λ, and g.

Answer 1

Concepts and reason

The concepts required to solve the given question are amplitude, frequency, and the velocity of the wave.

Initially, write an expression for the velocity of the wave. Later, calculate the angular frequency of the wave. Later, calculate the angular frequency. Finally, calculate the minimum amplitude of the wave.

Fundamentals

The expression for the velocity of the wave is as follows:

$v = \sqrt {\frac{{{T_{\rm{s}}}}}{\mu }}$

Here, ${T_{\rm{s}}}$ is the tension and $\mu$ is the mass per unit length.

The expression for the wave vector is as follows:

$k = \frac{{2\pi }}{\lambda }$

Here,$\lambda$ is the wavelength of the wave.

The expression for the angular frequency is as follows:

$\omega = vk$

Here, v is the velocity of the wave.

The expression for the minimum amplitude is as follows:

$A = \frac{g}{{{\omega ^2}}}$

Here, g is the acceleration due to gravity.

Substitute $\sqrt {\frac{{{T_{\rm{s}}}}}{\mu }}$ for v and $\frac{{2\pi }}{\lambda }$ for k in the equation$\omega = vk$.

$\omega = \left( {\sqrt {\frac{{{T_{\rm{s}}}}}{\mu }} } \right)\left( {\frac{{2\pi }}{\lambda }} \right)$

Substitute $\left( {\sqrt {\frac{{{T_{\rm{s}}}}}{\mu }} } \right)\left( {\frac{{2\pi }}{\lambda }} \right)$ for $\omega$ in the equation $A = \frac{g}{{{\omega ^2}}}$.

$\begin{array}{c}\\A = \frac{g}{{{{\left( {\left( {\sqrt {\frac{{{T_{\rm{s}}}}}{\mu }} } \right)\left( {\frac{{2\pi }}{\lambda }} \right)} \right)}^2}}}\\\\ = \frac{g}{{\frac{{{T_{\rm{s}}}}}{\mu }{{\left( {\frac{{2\pi }}{\lambda }} \right)}^2}}}\\\\ = \frac{{g\mu }}{{{T_{\rm{s}}}}}{\left( {\frac{\lambda }{{2\pi }}} \right)^2}\\\end{array}$

Ans:

The minimum amplitude of the wave is equal to $\frac{{g\mu }}{{{T_{\rm{s}}}}}{\left( {\frac{\lambda }{{2\pi }}} \right)^2}$ .