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Chapter 15: Problem 17

The upper end of a \(3.80-\) m-long steel wire is fastened to the ceiling, and a 54.0 -kg object is suspended from the lower end of the wire. You observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. What is the mass of the wire?

Short Answer

Expert verified
The mass of the wire is approximately 0.337 kg.

Step by step solution

01

Determine the speed of the pulse

To find the speed of the transverse pulse, use the formula for speed: \[ v = \frac{distance}{time} \]where the distance is the length of the wire, 3.80 m, and the time is 0.0492 s. Calculate \( v \):\[ v = \frac{3.80 \text{ m}}{0.0492 \text{ s}} \approx 77.24 \text{ m/s} \]
02

Relate pulse speed to tension and mass per unit length

The speed of a transverse wave on a string or wire is given by:\[ v = \sqrt{\frac{T}{\mu}} \]where \(T\) is the tension and \(\mu\) is the mass per unit length of the wire.
03

Calculate the tension in the wire

The tension in the wire is due to the weight of the suspended object. Thus, \[ T = mg \]where \( m = 54.0 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity). Calculate \( T \): \[ T = 54.0 \times 9.81 = 529.74 \text{ N} \]
04

Calculate mass per unit length

Use the relationship derived in Step 2 to calculate \( \mu \):\[ v = \sqrt{\frac{T}{\mu}} \rightarrow \mu = \frac{T}{v^2} \]Substitute the values of \(T\) and \(v\):\[ \mu = \frac{529.74}{(77.24)^2} \approx 0.0888 \text{ kg/m} \]
05

Calculate the mass of the wire

To find the total mass \(M\) of the wire, multiply the mass per unit length \(\mu\) by the total length of the wire:\[ M = \mu \times L \]where \(L = 3.80 \text{ m}\). Calculate \( M \):\[ M = 0.0888 \times 3.80 \approx 0.337 \text{ kg} \]

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Key Concepts

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

Transverse Wave
A transverse wave is a wave in which the oscillation is perpendicular to the direction the wave travels. Imagine a rope tied at one end and you create a wave by flicking the other end up and down; the wave moves horizontally along the rope, but the individual particles of the rope move up and down.

In the context of our exercise, the transverse pulse travels along the wire. Here, the wire's movement is side-to-side, while the pulse itself moves upward along its length from the object to the ceiling. Understanding how these waves travel is crucial in many fields, including physics and engineering.
Mass per Unit Length
The concept of mass per unit length, denoted as \( \mu \), is simply how much mass is contained in a segment of the wire of a given length. This is crucial because it helps us understand how heavy a segment of wire is without needing its full length or total mass.

In our example, this quantity \( \mu \) was calculated by using the formula \( \mu = \frac{T}{v^2} \), relating it to tension \( T \) and pulse speed \( v \). Knowing the mass per unit length allows us to determine the total mass of the wire by multiplying it by the wire's length.
Tension in a Wire
Tension in a wire refers to the force exerted by the wire when it is pulled tight by forces acting from opposite ends. In this exercise, the tension is a result of the gravitational force acting on the suspended mass.

We calculated the tension \( T \) using \( T = mg \), where \( m \) is the mass of the suspended object and \( g \) is the acceleration due to gravity. The tension is crucial because it affects the speed at which a wave travels through the wire; the greater the tension, the faster the wave can move.
Pulse Speed
Pulse speed refers to how fast a wave pulse travels along the wire. The formula to determine pulse speed is \( v = \frac{distance}{time} \).

In the exercise, we determined this speed by dividing the length of the wire by the time it took for the pulse to travel from the bottom to the top. The speed of the pulse is related to both the tension in the wire and the wire's mass per unit length as shown in the formula \( v = \sqrt{\frac{T}{\mu}} \). When you know two of these quantities, you can solve for the third.
Physics Problem-Solving
Physics problem-solving often involves breaking down complex phenomena into manageable steps, using established formulas and principles to understand what happens in a system.

With the example provided, the exercise started by solving simpler steps鈥攃alculating pulse speed, then the tension in the wire, followed by the mass per unit length, and finally, the wire's total mass.

Key problem-solving strategies include identifying known quantities, determining which physics laws and formulas apply, and using algebra to solve for unknowns. This approach can be applied in various physics problems to understand the core principles governing the system.

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Most popular questions from this chapter

A piano wire with mass 3.00 \(\mathrm{g}\) and length 80.0 \(\mathrm{cm}\) is stretched with a tension of 25.0 \(\mathrm{N}\) . A wave with frequency 120.0 \(\mathrm{Hz}\) and amplitude 1.6 \(\mathrm{mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

CALC Ant Joy Ride. You place your pet ant Klyde (mass \(m )\) on top of a horizontal, stretched rope, where he holds on tightly. The rope has mass \(M\) and length \(L\) and is under tension \(F .\) You start a sinusoidal transverse wave of wavelength \(\lambda\) and amplitude \(A\) propagating along the rope. The motion of the rope is in a vertical plane. Klyde's mass is so small that his presence has no effect on the propagation of the wave. (a) What is Klyde's top speed as he oscillates up and down? (b) Klyde enjoys the ride and begs for more. You decide to double his top speed by changing the tension while keeping the wavelength and amplitude the same. Should the tension be increased or decreased, and by what factor?

Weightless Ant. Ant. 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.

15.44\(\cdot\) The wave function of a standing wave is \(y(x, t)\) 4.44 \(\mathrm{mm} \sin [(32.5 \mathrm{rad} / \mathrm{m}) x] \sin [(754 \mathrm{rad} / \mathrm{s}) t] .\) For the two traveing waves that make up this standing wave, find the (a) amplitude (b) wavelength; (c) frequency; (d) wave speed; (e) wave function (f) From the information given, can you determine which hamonic this is? Explain.

A 1.50 -m string of weight 0.0125 \(\mathrm{N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W\) . Neglect 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 $$y(x, t)=(8.50 \mathrm{mm}) \cos \left(172 \mathrm{m}^{-1} x-4830 \mathrm{s}^{-1} t\right)$$ 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 wayes traveling down the string?
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