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Chapter 17: Problem 3

Two blocks each having a mass of \(3.2 \mathrm{~kg}\) are connected by a wire \(C D\) and the system is suspended from the ceiling by another wire \(A B\). The linear mass density of the wire \(A B\) is \(10 \mathrm{gm}^{-1}\) and that of \(C D\) is \(8 \mathrm{gm}^{-1}\). Find the speed of a transverse wave pulse produced in \(A B\) and in \(C D\).

Short Answer

Expert verified
The wave speeds are 79.21 m/s in wire AB and 88.51 m/s in wire CD.

Step by step solution

01

Convert Units

First, we need to convert the linear mass densities from grams per meter to kilograms per meter: For wire AB: \[10 \text{ gm/m} = 0.01 \text{ kg/m}\] For wire CD: \[8 \text{ gm/m} = 0.008 \text{ kg/m}\]
02

Calculate Tension in Wire CD

The tension in wire CD is determined by the weight of the two blocks. Since each block has a mass of 3.2 kg, the total mass is 6.4 kg. The force of gravity is given by:\[ T_{CD} = m \cdot g = 6.4 \times 9.8 = 62.72 \text{ N} \]
03

Calculate Speed of Wave in Wire CD

The speed of a wave in a wire is given by the formula:\[ v = \sqrt{\frac{T}{\mu}} \] For wire CD, using the already calculated tension and linear mass density:\[ v_{CD} = \sqrt{\frac{62.72}{0.008}} = \sqrt{7840} \approx 88.51 \text{ m/s}\]
04

Calculate Tension in Wire AB

The tension in wire AB supports both the weight of the blocks and the wire CD. Therefore, \[ T_{AB} = (6.4 \text{ kg} \times 9.8 \text{ m/s}^2) + (0.008 \text{ kg/m} \times \text{Length of CD in m} \times 9.8 \text{ m/s}^2) \]Assuming negligible length or that it's only due to the two blocks:\[ T_{AB} = 62.72 \text{ N} \]
05

Calculate Speed of Wave in Wire AB

Using the same wave speed formula for wire AB:\[ v_{AB} = \sqrt{\frac{62.72}{0.01}} = \sqrt{6272} \approx 79.21 \text{ m/s}\]
06

Conclusion

Therefore, the speed of the transverse wave pulse in wire AB is approximately 79.21 m/s, and in wire CD is approximately 88.51 m/s.

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Key Concepts

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

Tension in Wires
Understanding tension in wires is key to calculating wave speed. Tension is the force that stretches a wire. For example, if two blocks hang from a wire, the tension in that wire is the result of the gravitational force on the blocks. Tension is effectively the same as the weight of the objects, calculated as mass times gravity. In our example, two blocks, each weighing 3.2 kg, are attached to a wire, creating a total tension of 62.72 Newtons. This tension is a crucial factor in determining how waves travel along the wire.
Linear Mass Density
Linear mass density is a measure of how much mass exists in a unit length of a wire. It's usually measured in kilograms per meter. For wave calculations, linear mass density helps determine how fast waves can travel on a wire. If you know how much a given length of wire weighs, you can work out how dense it is. In the given exercise, for wire AB, the linear mass density is converted to 0.01 kg/m, and for wire CD, it is 0.008 kg/m. The lower the linear mass density, the faster a wave can move through the wire, as there is less mass to move.
Transverse Waves
Transverse waves travel perpendicular to the direction of the force applied. When a wave moves through a wire, it displaces parts of the wire up and down as it moves forward. This kind of wave can be seen in ropes or strings when they are plucked. In our example, transverse waves move through wires AB and CD. The speed of these waves depends on the balance between the tension in the wire and its linear mass density. By using the formula \( v = \sqrt{\frac{T}{\mu}} \), we can find the speed of transverse waves in different wires under various conditions.
Unit Conversion
Unit conversion is a fundamental step in physics to ensure consistency in calculations. Before solving any physics problem, make sure all your units are in the correct system, such as converting grams to kilograms for mass. In our problem, we convert the linear mass densities from grams per meter to kilograms per meter鈥�10 gm/m becomes 0.01 kg/m for wire AB and 8 gm/m becomes 0.008 kg/m for wire CD. This step ensures all measurements are compatible for solving the problem.
Physics Problem Solving
Successfully solving physics problems involves a systematic approach. Start by understanding the problem and identifying the required physics concepts. For wave speed calculation, pinpoint the tension in the wire and its linear mass density. Convert all units to the Standard International (SI) units. Then apply relevant formulas step by step. For instance, calculating wave speed using the formula \( v = \sqrt{\frac{T}{\mu}} \) becomes straightforward once tension and linear mass density are known. Always double-check your results to ensure their feasibility within the context of the problem.

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Most popular questions from this chapter

A \(200 \mathrm{~Hz}\) wave with amplitude \(1 \mathrm{~mm}\) travels on a long string of linear mass density \(6 \mathrm{~g} / \mathrm{m}\) kept under a tension of \(60 \mathrm{~N}\). (a) Find the average power transmitted across a given point on the string. (b) Find the total energy associated with the wave in a \(2.0 \mathrm{~m}\) long portion of the string.

A wave of frequency \(500 \mathrm{~Hz}\) has a wave velocity of \(350 \mathrm{~m} / \mathrm{s}\). (a) Find the distance between two points which are \(60^{\circ}\) out of phase. (b) Find the phase difference between two displacements at a certain point at time \(10^{-3} \mathrm{~s}\) apart.

You have learnt that a travelling wave in one dimension is represented by a function \(y=f(x, t)\) where \(x\) and \(t\) must appear in the combination \(a x \pm b t\) or \(x-v t\) or \(x+v t\), i.e. \(y=f(x \pm v t) .\) Is the converse true? Examine if the following functions for \(y\) can possibly represent a travelling wave (a) \((x-v t)^{2}\) (b) \(\log \left[(x+v t) / x_{0}\right]\) (c) \(1 /(x+v t)\)

Spherical waves are emitted from a \(1.0 \mathrm{~W}\) source in an isotropic non-absorbing medium. What the wave intensity \(1.0 \mathrm{~m}\) from the source?

The equation of a travelling wave is $$ y(x, t)=0.02 \sin \left(\frac{x}{0.05}+\frac{t}{0.01}\right) \mathrm{m} $$ Find (a) the wave velocity and (b) the particle velocity at \(x=0.2 \mathrm{~m}\) and \(t=0.3 \mathrm{~s}\). Given \(\cos \theta=-0.85\), where \(\theta=34 \mathrm{rad}\)
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