Q52P - University Physics with Modern Physics

Question

An ant with mass \(m\) is standing peacefully on top of a horizontal, stretched rope. The rope has mass per unit length \(\mu \) and is under tension \(F\).Without warning, Cousin Throckmorton starts a sinusoidal transverse wave of wave-length \(\lambda \) propagating along the rope. The motion of the rope is in a vertical plane. What minimum wave amplitude will make the ant become momentarily weightless? Assume that \(m\) is so small that the presence of the ant has no effect on the propagation of the wave.

Step-by-Step Solution

Answer

Thus, the minimum amplitude is \(2\;{\rm{mm}}\).

Step 1: Given in the question

Wavelength of each string \(\lambda  = 0.5\;{\rm{m}}\).

 

The mass per unit length \(\mu  = 0.1\;{\rm{kg/m}}\).

 

Tension is \(F = 3.125\;N\)

 

And gravity due to acceleration is \(g = {\pi ^2}\).

Step 2: Use formula of minimum amplitude

The maximum acceleration is written as \({a_{\max }} = {\omega ^2}A = g\). The angular velocity is \(\omega  = \frac{{2\pi v}}{\lambda }\). And the velocity is \(v = \sqrt {\frac{F}{\mu }} \). Thus, the minimum amplitude is given by:

 

\({A_{\min }} = \frac{{g{\lambda ^2}\mu }}{{4F}}\)

 

Here, \(F\) is tension, \(\mu \) is linear density .

Step 3: Calculate the minimum amplitude

According to the question,

 

\(\begin{array}{c}{A_{\min }} = \frac{{g{\lambda ^2}\mu }}{{4{\pi ^2}F}}\\ = \frac{{{\lambda ^2}\mu }}{{4F}}\\ = \frac{{0.5\;{\rm{m}} \times 0.5\;{\rm{m}} \times 0.1\;{\rm{kg/m}}}}{{3.125\;{\rm{N}}}}\\ = 2 \times {10^{ - 3}}\;{\rm{m}}\\ = 2\;{\rm{mm}}\end{array}\)

 

Hence, the minimum amplitude is \(2\;{\rm{mm}}\).