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Chapter 13: Problem 53

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 \(8.00 \mathrm{~N}\).

Short Answer

Expert verified
The required tension to produce transverse waves with a speed of 50.0 m/s on the string is 30.0 N (Newton). If the tension in the string is 8.00 N, the wave speed is 28.9 m/s.

Step by step solution

01

Determining the mass per unit length (linear density) of the string

The linear density \( \mu \) of a string is defined as its mass per unit length. It's given by the formula \( \mu = \frac{m}{L} \), where \(m\) is the total mass of the string and \(L\) is its length. In this case, \(m = 0.0600 \ kg \) and \( L = 5.00 \ m \). Substituting these values into the formula gives \( \mu = \frac{0.0600}{5.00} = 0.012 \ kg/m \).
02

Calculating the required tension in the string

The formula that relates the speed \(v\) of a wave on a string, the tension \(T\) in the string, and the string's linear density \( \mu \) is \( v = \sqrt{\frac{T}{\mu}} \). In this problem, the wave speed \( v = 50.0 \ m/s \) and \( \mu = 0.012 \ kg/m \) (from Step 1) are given, and the tension \( T \) is what we need to find. Re-arranging the formula for \( T \) gives \( T = \mu \times v^2 \). Substituting the given values into this formula gives \( T = 0.012 \ kg/m \times (50.0 \ m/s)^2 = 30.0 \ N \). So, the required tension in the string is 30.0 N.
03

Calculating the wave speed when the tension is 8.00 N

The final part of the problem requires us to find the wave speed when the tension is given as \( T = 8.00 \ N \). The same formula for wave speed we used in Step 2 applies here: \( v = \sqrt{\frac{T}{\mu}} \). Substituting the new tension and the linear density found in Step 1 into this equation gives \( v = \sqrt{\frac{8.00 \ N}{0.012 \ kg/m}} = 28.9 \ m/s \). So the wave speed when the tension is 8.00 N is 28.9 m/s.

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Most popular questions from this chapter

A spring oriented vertically is attached to a hard horizontal surface as in Figure P18.2. The spring has a force constant of \(1.46 \mathrm{kN} / \mathrm{m}\). How much is the spring compressed when a object of mass \(m=2.30 \mathrm{~kg}\) is placed on top of the spring and the system is at rest?

The period of motion of an object-spring system is \(T=\) \(0.528 \mathrm{~s}\) when an object of mass \(m=238 \mathrm{~g}\) is attached to the spring. Find (a) the frequency of motion in hertz and (b) the force constant of the spring. (c) If the total energy of the oscillating motion is \(0.234 \mathrm{~J}\), find the amplitude of the oscillations.

An object-spring system moving with simple harmonic motion has an amplitude \(A\). (a) What is the total energy of the system in terms of \(k\) and \(A\) only? (b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this situation, using only the variables for the mass \(m\), velocity \(v\), spring constant \(k\), and position \(x\). (c) Using the results of parts (a) and (b) and the conservation of energy equation, find the positions \(x\) of the object when its kinetic energy equals twice the potential energy stored in the spring. (The answer should in terms of \(A\) only.)

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