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With what tension must a rope with length \(2.50 \mathrm{~m}\) and mass \(0.120 \mathrm{~kg}\) be stretched for transverse waves of frequency \(40.0 \mathrm{~Hz}\) to have a wavelength of \(0.750 \mathrm{~m} ?\)

Short Answer

Expert verified
The rope must be stretched with a tension of \(43.2 \mathrm{N}\) to achieve the given conditions.

Step by step solution

01

Calculate wave speed

First, calculate the speed of the wave using the formula \(v = f\lambda\). Given a frequency of \(40.0 \mathrm{~Hz}\) and a wavelength of \(0.750 \mathrm{~m}\), the wave speed becomes \(v = 40.0 \mathrm{~Hz} \times 0.750 \mathrm{~m} = 30.0 \mathrm{~m/s}\).
02

Calculate linear density

Next, you need to calculate the linear density of the rope using the formula \(\mu = m / l\). Given a mass of \(0.120 \mathrm{~kg}\) and a length of \(2.50 \mathrm{~m}\), the linear density becomes \(\mu = 0.120 \mathrm{~kg} / 2.50 \mathrm{~m} = 0.048 \mathrm{~kg/m}\).
03

Calculate tension

Now that you have the wave speed \(v\) and the linear density \(\mu\), solve for the tension \(T\) in the string using the rearranged formula \(T = \mu v^2\). Substituting \(v = 30.0 \mathrm{~m/s}\) and \(\mu = 0.048 \mathrm{~kg/m}\), you get \(T = 0.048 \mathrm{~kg/m} \times (30.0 \mathrm{~m/s})^2 = 43.2 \mathrm{N}\).

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Key Concepts

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

Wave Speed Calculation
Understanding how to calculate the wave speed is essential for many physics problems, particularly those involving mechanical waves such as sound or waves on a rope, as in our exercise. Wave speed, denoted by the symbol 'v', is the speed at which a wave travels through a medium. The formula for calculating wave speed is a simple relationship: \( v = f\theta \).Here, 'f' represents the frequency of the wave, which is the number of oscillations or cycles the wave completes per second, and is measured in Hertz (Hz). 'f' simply tells you how often the wave repeats itself over time. 'f' is often given, as in the exercise where it was 40.0 Hz. On the other hand, '\theta’ denotes the wavelength, which is the distance between two consecutive similar points on the wave, such as peak to peak or trough to trough, and is measured in meters (m). In our example, this was 0.750 meters.By multiplying the frequency by the wavelength, we obtain the wave speed. The higher the frequency or the longer the wavelength, the faster the wave will travel. This relationship is useful to remember for problems that require you to solve for one of these quantities when the other two are known.
Linear Density
Linear density is a measure of mass per unit length of a material and is a critical property to consider when analyzing waves on strings or ropes. The linear density, typically symbolized by the Greek letter '\theta’, is determined using the formula:\( \theta = m / l \).In this equation, 'm' signifies the mass of the material, expressed in kilograms (kg), and 'l' is the length of the material in meters (m). For instance, if you have a rope with a mass of 0.120 kg and it is 2.50 m in length, as in our exercise, dividing the mass by the length yields a linear density of 0.048 kg/m. This property is directly related to the wave's properties on the rope. The lower the linear density, the easier it is for waves to travel along the rope, which, in turn, affects the wave speed as well as the tension required to produce a certain wave speed or frequency.
Wave Frequency and Wavelength Relationship
The relationship between wave frequency and wavelength is fundamental in understanding wave behavior. As briefly mentioned in the wave speed section, frequency, 'f', and wavelength, '\theta', are inversely related through the wave speed equation:\( v = f\theta \).This means when a wave's frequency increases, keeping the speed constant, the wavelength must decrease, and vice versa. The product of the frequency and wavelength always equals the speed of the wave for a given medium. This relationship also implies that if we know any two of these variables, we can easily solve for the third.When solving problems like the one from our exercise, where the goal is to determine the tension needed for a rope to carry a transverse wave of a certain frequency with a specific wavelength, the relationship between frequency and wavelength gives us the wave speed, which is a stepping-stone toward finding the tension. Remember, in a situation with a constant medium and tension, changing the frequency or wavelength of a wave will inherently affect the other if the wave speed is to remain unchanged.

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Most popular questions from this chapter

You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A \(2.00-\mathrm{cm}\) -long piece of the wire is left over from its installation. Using an analytical balance, you determine that the mass of the spare piece is \(14.5 \mu \mathrm{g}\). You then hang a \(0.400 \mathrm{~kg}\) mass from the lower end of the long, suspended wire. When a small-amplitude transverse wave pulse is sent up that wire, sensors at both ends measure that it takes the wave pulse \(26.7 \mathrm{~ms}\) to travel the length of the wire. (a) Use these measurements to calculate the length of the wire. Assume that the weight of the wire has a negligible effect on the speed of the transverse waves. (b) Discuss the accuracy of the approximation made in part (a).

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(M g\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu .\) You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M ;\) then you measure the node-to-node distance \(d\) for each standing-wave pattern and obtain the following data: $$ \begin{array}{l|lllll} \text { String } & \text { A } & \text { A } & \text { B } & \text { B } & \text { C } \\ \hline \mu(\mathrm{g} / \mathrm{cm}) & 0.0260 & 0.0260 & 0.0374 & 0.0374 & 0.0482 \\ M(\mathrm{~g}) & 559 & 249 & 365 & 207 & 262 \\ d(\mathrm{~cm}) & 48.1 & 31.9 & 32.0 & 24.2 & 23.8 \end{array} $$ (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^{2}(\) in \(\mathrm{kg} \cdot \mathrm{m})\) versus \(M(\) in \(\mathrm{kg}) .\) Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straightline fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). (d) For string A \((\mu=0.0260 \mathrm{~g} / \mathrm{cm}),\) what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of \(24.0 \mathrm{~cm}\) ? Use the value of \(f\) that you calculated in part (c).

One string of a certain musical instrument is \(75.0 \mathrm{~cm}\) long and has a mass of \(8.75 \mathrm{~g}\). It is being played in a room where the speed of sound is \(344 \mathrm{~m} / \mathrm{s}\). (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength \(0.765 \mathrm{~m} ?\) (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

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 traveling waves that make up this standing wave, find the (a) amplitude; (b) wavelength; (c) frequency; (d) wave speed; (e) wave functions. (f) From the information given, can you determine which harmonic this is? Explain.

For a violin, estimate the length of the portions of the strings that are free to vibrate. (a) The frequency of the note played by the open E5 string vibrating in its fundamental standing wave is 659 Hz. Use your estimate of the length to calculate the wave speed for the transverse waves on the string. (b) The vibrating string produces sound waves in air with the same frequency as that of the string. Use \(344 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air and calculate the wavelength of the E5 note in air. Which is larger: the wavelength on the string or the wavelength in air? (c) Repeat parts (a) and (b) for a bass viol, which is typically played by a person standing up. Start your calculation by estimating the length of the bass viol string that is free to vibrate. The G2 string produces a note with frequency \(98 \mathrm{~Hz}\) when vibrating in its fundamental standing wave.

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