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Chapter 13: Problem 56

A string is \(50.0 \mathrm{~cm}\) long and has a mass of \(3.00 \mathrm{~g}\). A wave travels at \(5.00 \mathrm{~m} / \mathrm{s}\) along this string. A second string has the same length, but half the mass of the first. If the two strings are under the same tension, what is the speed of a wave along the second string?

Short Answer

Expert verified
The speed of a wave along the second string is \(707 \mathrm{~m/s}\).

Step by step solution

01

Identify the Given Variables

In this step, identify and list the given variables: The length of the first string is \(50.0 \mathrm{~cm}\), its mass is \(3.00 \mathrm{~g}\), and the wave speed on it is \(5.00 \mathrm{~m/s}\). The second string has the same length but half the mass, and we're tasked to find the wave speed on the second string.
02

Calculate The Linear Density Of The First String

The formula to calculate the linear density (mass per unit length) of a string is \(\mu = \frac{m}{L} \), where \(\mu\) is the linear density, \(m\) is the mass, and \(L\) is the length. Substituting the given values, we find \(\mu_1 = \frac{3.00 \mathrm{~g}}{50.0 \mathrm{~cm}} = 0.06 \mathrm{~g/cm}\). Note that we're working in g/cm to make the calculation easier, but these units should be converted into SI units (kg/m) for final calculations.
03

Compute The Linear Density Of The Second String

As the second string has half the mass of the first, it's linear density is also half: \(\mu_2 = 0.5*\mu_1 = 0.5*0.06 \mathrm{~g/cm} = 0.03 \mathrm{~g/cm}\).
04

Use Wave Speed Formula

The wave speed (\(v\)) on a string is given by the formula \(v=\sqrt{\frac{T}{\mu}}\), where \(T\) is tension and \(\mu\) is the linear density. Because tension is the same for both strings, the ratio of their wave speeds will be inversely proportional to the square root of the ratio of their linear densities. Additionally, the wave speed on the first string is known, which allows to compute the wave speed on the second string: \(v_2 = v_1*\sqrt{\frac{\mu_1}{\mu_2}}\) . Substituting the known values, \(v_2 = 5.00 \mathrm{~m/s} * \sqrt{\frac{0.06 \mathrm{~g/cm}}{0.03 \mathrm{~g/cm}}} = 7.07 \mathrm{~m/s} \). After conversion to SI units, the result is \(v_2 = 707 \mathrm{~m/s}\) .
05

State the Final Answer

After performing the necessary calculations, the speed of the wave on the second string was found to be \(707 \mathrm{~m/s}\).

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Most popular questions from this chapter

A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes \(40.0\) vibrations in \(30.0 \mathrm{~s}\). Also, a given maximum travels \(425 \mathrm{~cm}\) along the rope in \(10.0 \mathrm{~s}\). What is the wavelength?

A vertical spring stretches \(3.9 \mathrm{~cm}\) when a \(10-\mathrm{g}\) object is hung from it. The object is replaced with a block of mass \(25 \mathrm{~g}\) that oscillates up and down in simple harmonic motion. Calculate the period of motion.

A bat can detect small objects, such as an insect, whose size is approximately equal to one wavelength of the sound the bat makes. If bats emit a chirp at a frequency of \(60.0 \times 10^{3} \mathrm{~Hz}\) and the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\), what is the smallest insect a bat can detect?

A system consists of a vertical spring with force constant \(k=1250 \mathrm{~N} / \mathrm{m}\), length \(L=1.50 \mathrm{~m}\), and object of mass \(m=5.00 \mathrm{~kg}\) attached to the end (Fig. P13.76). The object is placed at the level of the point of attachment with the spring unstretched, at position \(y_{i}=L\), and then it is released so that it swings like a pendulum. (a) Write Newton's second law symbolically for the system as the object passes through its lowest point. (Note that at the lowest point, \(r=\) \(L-y_{f}\).) (b) Write the conservation of energy equation symbolically, equating the total mechanical energies at the initial point and lowest point. (c) Find the coordinate position of the lowest point. (d) Will this pendulum's period be greater or less than the period of a simple pendulum with the same mass \(m\) and length \(L\) ? Explain.

Transverse waves with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\) are to be produced on a stretched string. A \(5.00-\mathrm{m}\) length of string with a total mass of \(0.0600 \mathrm{~kg}\) is used. (a) What is the required tension in the string? (b) Calculate the wave speed in the string if the tension is \(8.00 \mathrm{~N}\).
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