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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 wavelength \(\lambda\) 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.

Short Answer

Expert verified
The minimum amplitude is \(\frac{g \lambda^2 \mu}{4\pi^2 F}\).

Step by step solution

01

Understand the Condition for Weightlessness

The ant will become momentarily weightless when the acceleration of the wave's motion at the peak is equal to the acceleration due to gravity (\(g\)). This is because weightlessness occurs when the net apparent force acting on the ant is zero.
02

Express the Wave Motion Mathematically

A transverse wave on the rope can be expressed as:\[y(x,t) = A \sin\left(\frac{2\pi}{\lambda}x - \omega t\right)\]where \(A\) is the amplitude, \(\lambda\) is the wavelength, and \(\omega\) is the angular frequency.
03

Calculate the Maximum Vertical Acceleration

The vertical acceleration of the wave \(a\) can be found by taking the second partial derivative of the wave function with respect to time:\[a = \frac{\partial^2 y}{\partial t^2} = -A \omega^2 \sin\left(\frac{2\pi}{\lambda}x - \omega t\right)\]The maximum acceleration occurs when \(\sin(\cdot) = 1\):\[a_{max} = A \omega^2\]
04

Relate Maximum Acceleration to Gravity

For the ant to become momentarily weightless:\[a_{max} = g\]Substitute our expression for \(a_{max}\):\[A \omega^2 = g\]
05

Find Angular Frequency

The wave speed \(v\) is related to tension and mass per unit length by:\[v = \sqrt{\frac{F}{\mu}}\]Angular frequency is related to wave speed and wavelength by:\[\omega = \frac{2\pi v}{\lambda}\]
06

Substitute Angular Frequency into the Condition

Substitute \(\omega\) into the condition for weightlessness:\[A \left(\frac{2\pi v}{\lambda}\right)^2 = g\]Solve for \(A\):\[A = \frac{g \lambda^2}{4\pi^2 v^2}\]
07

Express Wave Speed in Terms of Known Quantities

Substitute the expression for \(v\):\[A = \frac{g \lambda^2}{4\pi^2 \frac{F}{\mu}}\] Simplify to:\[A = \frac{g \lambda^2 \mu}{4\pi^2 F}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Motion Equations
In a sinusoidal transverse wave, the displacement of any point on the rope can be described by a wave motion equation. This equation shows how the height (often called the amplitude) of the wave varies as a function of both position and time. The basic form of a wave motion equation for a transverse wave propagating along a rope is given by:\[y(x,t) = A \sin\left(\frac{2\pi}{\lambda}x - \omega t\right)\]Here, \(y(x,t)\) is the displacement of the wave, \(A\) is the amplitude, \(\lambda\) is the wavelength, \(x\) is position, \(t\) is time, and \(\omega\) is the angular frequency.
  • The amplitude \(A\) represents the maximum vertical displacement from the equilibrium position.
  • The wavelength \(\lambda\) is the distance over which the wave's shape repeats.
  • The angular frequency \(\omega\) determines how fast the wave oscillates over time.
This equation is fundamental in understanding how waves transmit energy through a medium, like a rope, without causing permanent displacement.
Vertical Acceleration
Vertical acceleration in wave motion refers to how quickly the wave's height changes vertically over time. To find this, we need to look at the second derivative of the displacement function with respect to time. The resulting formula for vertical acceleration is:\[a = \frac{\partial^2 y}{\partial t^2} = -A \omega^2 \sin\left(\frac{2\pi}{\lambda}x - \omega t\right)\]The minus sign indicates that the acceleration is in the opposite direction of the displacement, exhibiting a restorative nature.
  • At maximum acceleration, the sine part of the equation equals 1, meaning it's pulling down with full strength, which would be equal to \(A \omega^2\).
  • When this matches the gravitational acceleration \(g\), any object, like our ant, may become momentarily weightless if on the wave's peak.
Understanding this concept helps in analyzing how wave mechanics potentially affect physical objects.
Angular Frequency
Angular frequency, \(\omega\), is a critical concept in wave analysis reflecting how quickly a wave oscillates over time. Unlike simple frequency which is cycles per second, angular frequency refers to how many radians per second the wave covers.Calculated as:\[\omega = \frac{2\pi v}{\lambda}\]This formula ties together wave speed \(v\) and wavelength \(\lambda\).
  • Angular frequency helps determine the sinusoidal properties of the wave, dictating how "stretched out" or "condensed" the wave appears over time.
  • It is closely related to the periodicity of waves: higher angular frequency indicates faster oscillations.
Grasping the concept of angular frequency is integral to mastering the description and prediction of wave behaviors.
Wave Speed Calculation
The speed at which a wave travels through a medium like a rope can be found by considering the forces acting on the rope. The wave speed \(v\) is connected to the tension \(F\) in the rope and the mass per unit length \(\mu\):\[v = \sqrt{\frac{F}{\mu}}\]This formula shows:
  • Wave speed increases with greater tension because tighter ropes allow waves to propagate faster.
  • A lower mass per unit length \(\mu\) also results in higher wave speed, as the rope is lighter and easier to move.
Combining this with the wavelength, we can calculate the angular frequency and further analyze wave behavior. This understanding allows for practical applications in engineering and physics, such as ensuring the stability and integrity of structures that utilize cable systems.

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Most popular questions from this chapter

A transverse wave on a rope is given by $$y(x, t) = (0.750 \, \mathrm{cm}) \mathrm{cos} \space \pi[(10.400 \, \mathrm{cm}^{-1})x + (250 \mathrm s^{-1})t]$$ (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at these values of \(t:\) 0, 0.0005 s, 0.0010 s. (c) Is the wave traveling in the \(+x-\) or \(-x\)-direction? (d) The mass per unit length of the rope is 0.0500 kg/m. Find the tension. (e) Find the average power of this wave.

One string of a certain musical instrument is 75.0 cm long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 m/s. (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 m? (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t) = (8.50 \, \mathrm{mm}) \mathrm{cos}(172 \, \mathrm{rad/m} \space x - 4830 \, \mathrm{rad/s} \space t)$$ Assume that the tension of the string is constant and equal to \(W\). (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W\)? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling \(down\) the string?

A 1.50-m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 62.0 m/s. What are the wavelength and frequency of (a) the fundamental; (b) the second overtone; (c) the fourth harmonic?

A wave on a string is described by \(y(x, t) = A \mathrm{cos}(kx - \omega t)\). (a) Graph \(y, v_y\), and \(a_y\) as functions of \(x\) for time \(t = 0\). (b) Consider the following points on the string: (i) \(x =\) 0; (ii) \(x = \pi/4k\); (iii) \(x = \pi/2k\); (iv) \(x = 3\pi/4k\); (v) \(x = \pi k\); (vi) \(x = 5\pi/4k\); (vii) \(x = 3\pi/2k\); (viii) \(x = 7\pi/4k\). For a particle at each of these points at \(t = 0\), describe in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not accelerating.

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