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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 Hz. The other end passes over a pulley and supports a 1.50-kg mass. The linear mass density of the rope is 0.0480 kg/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 kg?

Short Answer

Expert verified
(a) 17.46 m/s; (b) 0.1455 m; increasing mass to 3.00 kg gives (c) 24.67 m/s and 0.2056 m respectively.

Step by step solution

01

Understanding Wave Speed on a String

The wave speed \( v \) on a string is determined by the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string and \( \mu \) is the linear mass density. The tension \( T \) is the weight of the mass and is given by \( T = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately 9.8 \( m/s^2 \)).
02

Calculating Tension with 1.50 kg Mass

Calculate the tension in the rope when the mass is 1.50 kg. Using \( T = mg \), we find \( T = (1.50 \, \text{kg}) \times (9.8 \, \text{m/s}^2) = 14.7 \, \text{N} \).
03

Calculating Wave Speed with 1.50 kg Mass

Substitute the values of \( T \) (14.7 N) and \( \mu \) (0.0480 kg/m) into the wave speed formula: \( v = \sqrt{\frac{14.7}{0.0480}} \). This results in a wave speed \( v \approx 17.46 \, \text{m/s} \).
04

Calculating Wavelength for 1.50 kg Mass

The wavelength \( \lambda \) can be determined from the relationship \( v = f \lambda \), where \( f = 120 \, \text{Hz} \). Using \( v = 17.46 \, \text{m/s} \), solve for \( \lambda \): \( \lambda = \frac{v}{f} = \frac{17.46}{120} \approx 0.1455 \, \text{m} \).
05

Calculating Wave Speed with 3.00 kg Mass

Now increase the mass to 3.00 kg. Calculate the tension: \( T = (3.00 \, \text{kg}) \times (9.8 \, \text{m/s}^2) = 29.4 \, \text{N} \). Then the wave speed is \( v = \sqrt{\frac{29.4}{0.0480}} \approx 24.67 \, \text{m/s} \).
06

Calculating Wavelength for 3.00 kg Mass

Using the new wave speed \( v = 24.67 \, \text{m/s} \), calculate the new wavelength: \( \lambda = \frac{v}{f} = \frac{24.67}{120} \approx 0.2056 \, \text{m} \).

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Key Concepts

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

Transverse Waves
Transverse waves are a type of wave where the motion of the medium is perpendicular to the direction of the wave. In the case of a rope being vibrated by a tuning fork, the rope moves up and down, while the wave travels horizontally along the rope. These waves occur frequently in various forms, such as waves on a string or ripples on the surface of water.
Understanding transverse waves is crucial as it helps us analyze how energy is transferred through these mediums without the medium itself moving far from its original position. This concept applies not just to physical ropes but also to electromagnetic waves, where electric and magnetic fields oscillate perpendicular to the direction of wave travel.
For students working through the exercise, visualizing the up-and-down movement of the rope while picturing the wave moving forward can clarify the transverse nature of these waves.
Wave Speed
Wave speed is how fast a wave travels through a medium. For transverse waves on a string, the wave speed can be calculated using the formula: \[ v = \sqrt{\frac{T}{\mu}} \]where \( v \) is the wave speed, \( T \) is the tension in the string, and \( \mu \) is the linear mass density of the string.
This formula illustrates that the speed of the wave is influenced by both the tension and the mass density of the string.
  • If the tension is higher, the wave speed increases. Think of pulling a string very tight - vibrations move faster because there's more force along the string.
  • a higher mass density (meaning the string is "heavier" for its length) slows down the wave, much like pushing a denser object takes more energy.
Understanding these factors helps in predicting and explaining the behavior of different waves in physics and in practical scenarios.
Wavelength
Wavelength refers to the distance between two consecutive crests or troughs of a wave. It represents one complete cycle of the wave. For the exercise, the wavelength was calculated using the equation:\[ \lambda = \frac{v}{f} \]where \( \lambda \) is the wavelength, \( v \) is the wave speed, and \( f \) is the frequency.
The frequency in this context is how often the wave cycles occur per second, measured in hertz (Hz). A higher wave speed or a lower frequency results in a longer wavelength.
  • This means with a faster or less frequent wave, the distance between peaks will increase.
  • Conversely, if a wave is traveling slower or cycling rapidly, the wavelength shortens.
When calculating, never forget that any change in wave speed or frequency impacts the wavelength directly, showing the dynamic nature of waves in physics.
Tension in Strings
Tension in strings is a critical factor influencing wave mechanics. It is the force exerted along the string when it is pulled tight. In the exercise, the tension \( T \) is calculated using:\[ T = mg \]where \( m \) is the mass hanging from the string and \( g \) is the acceleration due to gravity \((9.8 \, \text{m/s}^2)\).
This tension is what allows the wave to travel through the string. Larger masses increase the tension, leading to faster waves because the stronger force makes the string more responsive to vibrational movements. Conversely, less tension slows the wave. Consider:
  • A dramatically tighter string (more tension) supports faster traveling waves, like a drum, which is taut so it can produce sound efficiently.
  • If the string is loose, the wave speed decreases, leading to a less "energetic" system.
Understanding tension and its impact on wave speed is essential for studying mechanical waves and their applications in areas from music to engineering.

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Most popular questions from this chapter

A transverse wave on a rope is given by $$y(x, t) = (0.750 \, \mathrm{cm}) \mathrm{cos} \space \pi[(10.400 \, \mathrm{cm}^{-1})x + (250 \mathrm s^{-1})t]$$ (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at these values of \(t:\) 0, 0.0005 s, 0.0010 s. (c) Is the wave traveling in the \(+x-\) or \(-x\)-direction? (d) The mass per unit length of the rope is 0.0500 kg/m. Find the tension. (e) Find the average power of this wave.

A 1750-N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (\(A\) and \(B\)), each 1.25 m long and weighing 0.290 N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached. If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling? Which pulse arrives first? (Ignore the effect of the weight of the wires on the tension in the wires.)

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 \(Mg\) 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 standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^2\) (in kg \(\cdot\) m) versus \(M\) (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g = 9.80 \, \mathrm{m/s}^2\). (d) For string A (\(\mu = 0.0260\) g/cm), what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of \(f\) that you calculated in part (c).

By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is 0.026 W/m\(^2\) at a distance of 4.3 m from the source. (a) What is the intensity at a distance of 3.1 m from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

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