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One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at \(120 \mathrm{~Hz}\). The other end passes over a pulley and supports a \(1.50 \mathrm{~kg}\) mass. The linear mass density of the rope is \(0.0480 \mathrm{~kg} / \mathrm{m} .\) (a) What is the speed of a transverse wave on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change if the mass were increased to \(3.00 \mathrm{~kg} ?\)

Short Answer

Expert verified
For the initial situation, solve the equations in steps 1 and 2. Then, when the mass is increased to 3.00 kg, calculate the new tension, recalculate the speed and wavelength of the wave using steps 1 and 2.

Step by step solution

01

Calculate the speed of wave

First, calculate the speed (v) of the transverse wave. The speed of a wave in a string depends on the square root of the tension (\(T\)) in the string divided by the linear mass density (\(饾渿\)) of the string. For part (a), the tension is caused by the 1.50 kg weight, which equals mass (m) times gravity (g) or \(1.50 kg * 9.8 m/s^2 = 14.7 N\). So, the equation for \(v\) becomes: \(v = \sqrt{\frac{T}{饾渿}} = \sqrt{\frac{14.7 N}{0.0480 kg/m}}\). Solve for \(v\)
02

Calculate the wavelength

Next, for part (b), calculate the wavelength (位) of the transverse wave. The wavelength is equal to the speed of the wave divided by frequency (f). So the equation for 位 is: 位 = \(\frac{v}{f} = \frac{v}{120 Hz}\). Solve for 位
03

Re-calculation for increased mass

Finally, for part (c), re-calculate the speed and wavelength of the wave when the mass is increased to 3.00 kg. This will increase the tension in the string to \(3.00 kg * 9.8 m/s^2 = 29.4 N\). Use this new tension to repeat the calculations in steps 1 and 2.

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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 speed of a transverse wave in a string is a fundamental aspect of wave physics. The general formula used to determine the speed of a wave on a string is given by the square root of the ratio of the tension in the string to the linear mass density, expressed mathematically as:
\[ v = \sqrt{\frac{T}{\text{饾渿}}} \]
where \( v \) is the speed of the wave, \( T \) is the tension in the string, and \( 饾渿 \) (Greek letter mu) represents the linear mass density of the string. In the given exercise, the tension is produced by a hanging mass which applies a force due to gravity. To find the tension, you multiply the mass \( m \) by the acceleration due to gravity \( g \), or \( T = m \times g \).To visualize, consider a guitar string. When plucked, the tension in the string combined with how light or heavy the string is (its linear mass density) determines how fast the vibration travels, creating sound at a specific pitch. The same concept applies to any transverse wave traveling along a medium, where the medium's properties influence the speed of the wave.
Tuning Fork Vibration
A tuning fork is a simple device used to produce a fixed tone at a precise frequency, which is determined by the fork's size and material. When the prong of a tuning fork vibrates, it sets the surrounding medium into vibrational motion, creating a sound wave with the same frequency as the fork. In our problem, the tuning fork vibrates at a frequency of \( 120 \) Hz, meaning that it completes 120 oscillations or cycles per second.

This frequency directly correlates with the number of transverse waves produced in the rope per second. With this information, you can then determine the wavelength of the waves in the rope by looking at the wave speed and frequency relation. A practical application of this concept is in musical instruments, where adjusting the tension in the strings changes the pitch (frequency) of the sound produced.
Linear Mass Density
Linear mass density, often denoted by \( 饾渿 \), is a measure of mass per unit length of a material, such as a string or a rope. It's given in units of mass over distance (kg/m). This property impacts how waves propagate along the medium; the higher the linear mass density, the slower a wave moves for a given tension.

For example, consider different types of strings on a musical instrument: a thicker string (which has a higher linear mass density) will produce a wave that travels slower compared to a thinner string, assuming the tension is the same. In our scenario, the rope has a linear mass density of \( 0.0480 \) kg/m, which means for every meter of rope, there's 0.0480 kilograms of mass. Adjusting the linear mass density, much like tension, alters the properties of the wave speed and subsequently affects the wavelength and frequency of waves in the medium.
Wave Frequency and Wavelength
Wave frequency and wavelength are intrinsically linked properties of waves. The frequency (\( f \)) of a wave is the number of oscillations that pass a point per unit of time, measured in hertz (Hz). The wavelength (\( \lambda \)) is the physical length of one complete wave cycle, measured in meters.

The relationship between wave speed (\( v \)), frequency, and wavelength is defined by the equation:
\[ v = f \times \lambda \]
In simpler terms, wave speed is equal to how frequently the waves pass (the frequency) times the length of one wave (wavelength). So, if you know the speed of the wave and its frequency 鈥 as you do in our exercise where the tuning fork vibrates at 120 Hz 鈥 you can calculate the wavelength. As you might witness in ocean waves, where each wave can be meters apart (long wavelength) and follow each other in succession (frequency), these principles of wave frequency and wavelength apply across different contexts, from the microscopic to the macroscopic.

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Most popular questions from this chapter

Tsunami! On December \(26,2004,\) a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed more than 200,000 people. Satellites observing these waves from space measured \(800 \mathrm{~km}\) from one wave crest to the next and a period between waves of 1.0 hour. What was the speed of these waves in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) ? Does your answer help you understand why the waves caused such devastation?

A certain transverse wave is described by $$ y(x, t)=(6.50 \mathrm{~mm}) \cos 2 \pi\left(\frac{x}{28.0 \mathrm{~cm}}-\frac{t}{0.0360 \mathrm{~s}}\right) $$ Determine the wave's (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e) direction of propagation.

A small bead of mass \(4.00 \mathrm{~g}\) is attached to a horizontal string. Transverse waves of amplitude \(A=0.800 \mathrm{~cm}\) and frequency \(f=20.0 \mathrm{~Hz}\) are set up on the string. Assume the mass of the bead is small enough that the bead doesn't alter the wave motion. During the wave motion, what is the maximum vertical force that the string exerts on the bead?

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).

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.

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