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A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of \(240 \mathrm{m} / \mathrm{s} .\) What would be the speed of the wave on the longer wire?

Short Answer

Expert verified
The wave speed on the longer wire is approximately 339.36 m/s.

Step by step solution

01

Identify the wave speed formula

Recall the formula for wave speed on a stretched string, which is given by \( v = \sqrt{\frac{T}{\mu}} \), where \(v\) is the wave speed, \(T\) is the tension in the wire, and \(\mu\) is the linear mass density (mass per unit length).
02

Determine the relationship between mass and length

Since both wires have the same mass \(m\) and the tension \(T\) is the same, we can see that the linear density for each wire is \(\mu = \frac{m}{L}\), where \(L\) is the length of the wire. For the longer wire, the length is twice that of the shorter wire.
03

Compare the lengths of the wires

Let \(L_s\) be the length of the shorter wire and \(L_l\) be the length of the longer wire. Since \(L_l = 2L_s\), the linear mass density for the shorter wire is \(\mu_s = \frac{m}{L_s}\), and for the longer wire \(\mu_l = \frac{m}{2L_s}\).
04

Apply the wave speed formula to the shorter wire

For the shorter wire, the wave speed is given as \(240 \ \mathrm{m/s}\). Therefore, \(240 = \sqrt{\frac{T}{\frac{m}{L_s}}} = \sqrt{\frac{TL_s}{m}}\).
05

Apply the wave speed formula to the longer wire

For the longer wire, the speed \(v_l\) is \(v_l = \sqrt{\frac{T}{\frac{m}{2L_s}}} = \sqrt{\frac{2TL_s}{m}}\).
06

Relate the speeds on both wires

We see that the speed of the wave on the longer wire becomes \(v_l = \sqrt{2} \times \sqrt{\frac{TL_s}{m}}\). Since \(\sqrt{\frac{TL_s}{m}} = 240\), substitute this to get \(v_l = \sqrt{2} \times 240\).
07

Calculate the wave speed on the longer wire

Calculate \(v_l = \sqrt{2} \times 240 = 240\sqrt{2}\). By approximating \(\sqrt{2} \approx 1.414\), \(v_l \approx 1.414 \times 240 \approx 339.36 \ \mathrm{m/s}\).

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Key Concepts

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

Tension in Wire
In understanding wave speed on a string or wire, a crucial factor is the tension in the wire. Tension refers to the force exerted along the wire, which is responsible for stretching it tight. Imagine a strong pull on both ends of the wire, keeping it taut and ready for a wave to travel through. The greater this tension, the faster waves can propagate across it. This is because higher tension means that the material is stiffer, allowing the wave's energy to transfer more efficiently. This principle can be described by the formula: the wave speed is proportional to the square root of the tension divided by the linear mass density.
Linear Mass Density
The concept of linear mass density, denoted by \( \mu \), is a measure of how much mass is distributed along the length of a wire. It is calculated by dividing the total mass of the wire by its length, \( \mu = \frac{m}{L} \). Think of it as how tightly packed the mass of the wire is along its span. A wire with more mass packed into a shorter length will have a higher linear mass density. This is significant because linear mass density appears in the wave speed formula: \( v = \sqrt{\frac{T}{\mu}} \). When the wire is stretched longer but retains the same mass, its linear mass density decreases. This affects the speed at which waves travel on it, as seen in the exercise where the longer wire has a reduced linear mass density.
Transverse Waves
Transverse waves are a type of wave where the disturbance moves perpendicular to the direction of the wave's travel. Imagine wiggling a rope up and down; the wave travels horizontally along the rope as your hand creates motion vertically. This up-and-down motion is characteristic of transverse waves. On a wire, transverse waves travel as the wire itself oscillates in a direction perpendicular to its length. Understanding this type of motion is key to analyzing wave behaviors and calculating various properties like speed. These waves play a vital role as they travel through the medium, influenced by tension and linear mass density.
Wave Velocity in Strings
Wave velocity in strings is a fundamental aspect when studying waves on a wire. Wave velocity, or wave speed, is how fast a wave travels along the string or wire. For strings under tension, this velocity depends on both the tension of the string and its linear mass density. The mathematical relationship is given by \( v = \sqrt{\frac{T}{\mu}} \), where \(v\) is the wave velocity, \(T\) is the tension, and \(\mu\) is the linear mass density. This equation shows us that:
  • Increased tension leads to increased wave speed.
  • Lower linear mass density results in higher wave speed.
In the exercise, we see that when the wire is lengthened, the linear mass density decreases, which under constant tension results in a faster wave speed on the longer wire. This principle ensures waves travel efficiently across materials with specific tension and density characteristics.

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Most popular questions from this chapter

A car is parked \(20.0 \mathrm{m}\) directly south of a railroad crossing. A train is approaching the crossing from the west, headed directly east at a speed of \(55.0 \mathrm{m} / \mathrm{s}\). The train sounds a short blast of its \(289-\mathrm{Hz}\). horn when it reaches a point \(20.0 \mathrm{m}\) west of the crossing. What frequency does the car's driver hear when the horn blast reaches the car? The speed of sound in air is \(343 \mathrm{m} / \mathrm{s} .\)

A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than to the other. He fires a gun and, after a while, hears three echoes. The second echo arrives \(1.6 \mathrm{s}\) after the first, and the third echo arrives \(1.1 \mathrm{s}\) after the second. Assuming that the speed of sound is \(343 \mathrm{m} / \mathrm{s}\) and that there are no reflections of sound from the ground, find the distance between the cliffs.

A Mysterious Underwater Object. You and your team are on a reconnaissance mission in a submarine exploring a mysterious object in the cold waters of the Weddell Sea, off the coast of Antarctica. The sonar indicates that the object, which had otherwise been moving erratically, has changed course and is now on a direct collision course with your sub. The captain issues an "all stop" order, bringing the sub to a halt relative to the water. The sonar operator "pings" the object, which amounts to sending a short blast of sound in the direction of the object. The emitted sound wave has a frequency of \(1550 \mathrm{Hz}\) and a speed of \(1552 \mathrm{m} / \mathrm{s}\) (the speed of sound in seawater). The sound reflects from the object and returns \(2.582 \mathrm{s}\) after it was emitted from your sub, and its frequency has shifted to \(1598 \mathrm{Hz}\). (a) How far from the sub was the object when the sound reflected from it? (b) What is the object's speed? (c) How long after you receive the return signal will it take the object to reach your submarine?

An observer stands \(25 \mathrm{m}\) behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of the bullets is \(840 \mathrm{m} / \mathrm{s},\) and the air temperature is \(20^{\circ} \mathrm{C} .\) How far does each bullet travel before the observer hears the report of the rifle? Assume that the bullets encounter no obstacles during this interval, and ignore both air resistance and the vertical component of the bullets" motion.

The mass of a string is \(5.0 \times 10^{-3} \mathrm{kg}\), and it is stretched so that the tension in it is \(180 \mathrm{N}\). A transverse wave traveling on this string has a frequency of \(260 \mathrm{Hz}\) and a wavelength of \(0.60 \mathrm{m} .\) What is the length of the string?

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