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Adjacent antinodes of a standing wave on a string are 15.0 \(\mathrm{cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 \(\mathrm{cm}\) and period 0.0750 \(\mathrm{s}\) . The string lies along the \(+x\) -axis and is fixed at \(x=0 .(\text { a) How far }\) apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

Short Answer

Expert verified
(a) 7.5 cm apart. (b) Wavelength is 30.0 cm, amplitude is 0.850 cm, speed is 400 cm/s. (c) Max speed 71.21 cm/s, min speed 0 cm/s. (d) 7.5 cm.

Step by step solution

01

Calculate Distance between Nodes

For part (a), the distance between adjacent antinodes is given as 15.0 cm. Since the distance between adjacent nodes is half of the distance between adjacent antinodes in a standing wave, the distance between adjacent nodes is calculated as:\[ \text{Distance between nodes} = \frac{15.0}{2} \text{ cm} = 7.5 \text{ cm} \]
02

Determine Wavelength of Traveling Waves

In part (b), the wavelength \( \lambda \) is twice the distance between successive antinodes. Therefore, the wavelength is:\[ \lambda = 2 \times 15.0 \text{ cm} = 30.0 \text{ cm} \]
03

Calculate Amplitude of Traveling Waves

The amplitude of the individual traveling waves is the same as the given amplitude of the displacement at an antinode in the standing wave, which is 0.850 cm.
04

Calculate Speed of Traveling Waves

To calculate the speed \( v \) of the traveling waves, use the formula:\[ v = \frac{\lambda}{T} \]where \( T \) is the period of oscillation (0.0750 s). Substituting the known values:\[ v = \frac{30.0 \text{ cm}}{0.0750 \text{ s}} = 400 \text{ cm/s} \]
05

Calculate Maximum Transverse Speed at an Antinode

For part (c), the maximum transverse speed \( v_{max} \) at an antinode is given by:\[ v_{max} = A \omega \]where \( A \) is the amplitude (0.850 cm) and \( \omega \) is the angular frequency, calculated as \( \omega = \frac{2\pi}{T} \). Thus:\[ \omega = \frac{2\pi}{0.0750} = \frac{2\pi}{0.0750} \approx 83.78 \text{ rad/s} \]Then:\[ v_{max} = 0.850 \times 83.78 \approx 71.21 \text{ cm/s} \]
06

Calculate Minimum Transverse Speed at an Antinode

The minimum transverse speed at an antinode is 0 cm/s, as the speed becomes zero at the maximum displacement point.
07

Determine Shortest Distance from Node to Antinode

In part (d), the shortest distance between a node and an antinode is a quarter of the wavelength:\[ \text{Shortest distance} = \frac{\lambda}{4} = \frac{30.0}{4} = 7.5 \text{ cm} \]

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Key Concepts

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

Nodes and Antinodes
Standing waves are fascinating phenomena in physics where certain points along the medium, called nodes, remain stationary, while adjacent points called antinodes experience the maximum oscillation. In simpler terms, nodes are points of no movement, whereas antinodes are points where the movement is greatest.
To visualize this, imagine a jump rope fixed at both ends. When you wave one end just right, you'll create points where the rope doesn't move up or down - these are nodes. Meanwhile, the peaks and valleys at the other points are the antinodes.
In the given exercise, the distance between adjacent antinodes is 15.0 cm. Nodes occur midway between antinodes, so the distance between nodes is half that of antinodes, which is 7.5 cm.
Wave Parameters
Wave parameters define the characteristics of a wave, including amplitude, wavelength, velocity, and frequency. Amplitude is the height from the equilibrium position to a peak, describing how much energy the wave carries.
The exercise provides an amplitude of 0.850 cm, illustrating how far a point at an antinode moves from the center position.
  • Amplitude: 0.850 cm
  • Period: 0.0750 s
  • Wavelength: Calculated as 30.0 cm
  • Speed: Found to be 400 cm/s
The amplitude of the individual traveling waves that form the standing wave is the same as the amplitude of the standing wave.
Simple Harmonic Motion
Simple harmonic motion (SHM) describes the type of periodic motion that occurs when the net force acting on a mass is directly proportional to its displacement, with the force always directed toward the equilibrium position.
This is the same motion exhibited by a particle at an antinode on a standing wave. The oscillation of this particle is characterized by its amplitude and period.
In SHM, the maximum speed of a particle can be determined by the product of the amplitude and angular frequency, where angular frequency, \( \omega \), can be calculated using:\[ \omega = \frac{2\pi}{T} \]This exercise calculates the maximum speed at an antinode using the given values and the formula for ¥(v_{max}=Aω).
Wave Speed
Wave speed indicates how fast the wave energy is traveling through the medium. It is influenced by the medium's properties, such as tension and density in the case of waves on a string.
The speed is computed using the formula:\[ v = \frac{\lambda}{T} \]
where \( \lambda \) is the wavelength and \( T \) is the period of the wave. From the values extracted earlier, the wave speed for the traveling waves in this scenario is calculated as 400 cm/s, showcasing how fast the wave's energy propagates along the string.
Understanding wave speed is crucial as it helps link the physical properties of a medium to the dynamic characteristics of the wave.
Wavelength
Wavelength is the spatial period of a wave—the distance over which the wave's shape repeats. This can be the distance between consecutive antinodes, nodes, or any two points that are in phase.
For a standing wave, the wavelength is twice the distance between adjacent antinodes, which is calculated as 30.0 cm in the exercise. This measure is vital as it determines the wave's capacity to interfere constructively or destructively as well as its influence in various physical and acoustic phenomena.
In this exercise, each feature like nodes, antinodes, and wavelength comes together to describe the standing wave completely, helping deepen the understanding of wave behavior.

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Most popular questions from this chapter

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 0.0550 \(\mathrm{kg} / \mathrm{m}\) . (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 \(\mathrm{kg}\) ?

Ant Joy Ride. You place your pet ant Klyde (mass \(m )\) on top of a horizontal, stretched rope, where he holds on tightly. The rope has mass \(M\) and length \(L\) and is under tension \(F\) . You start a sinusoidal transverse wave of wavelength \(\lambda\) and amplitude \(A\) propagating along the rope. The motion of the rope is in a vertical plane. Klyde's mass is so small that his presence has no effect on the propagation of the wave. (a) What is Klyde's top speed as he oscillates up and down? (b) Klyde enjoys the ride and begs for more. You decide to double his top speed by changing the tension while keeping the wavelength and amplitude the same. Should the tension be increased or decreased, and by what factor?

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t)=\) \((5.60 \mathrm{cm}) \sin\) \([(0.0340 \mathrm{rad} / \mathrm{cm}) x]\) sin \([(50.0 \mathrm{rad} / \mathrm{s}) t]\) where the origin is at the left end of the string, the \(x\) -axis is along the string and the \(y\) -axis is perpendicular to the string. (a) Draw a sketch that shows the standing wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x, t)\) for this string if it were vibrating in its eighth harmonic?

Waves on a Stick. A flexible stick 2.0 \(\mathrm{m}\) long is not fixed in any way and is free to vibrate. Make clear drawings of this stick vibrating in its first three harmonics, and then use your drawings to find the wavelengths of each of these harmonics. (Hint: Should the ends be nodes or antinodes?)

The B-string of a guitar is made of steel (density \(\left.7800 \mathrm{~kg} / \mathrm{m}^{3}\right),\) is \(63.5 \mathrm{~cm}\) long, and has diameter \(0.406 \mathrm{~mm}\) The fundamental frequency is \(f=247.0 \mathrm{~Hz}\) (a) Find the string tension. (b) If the tension \(F\) is changed by a small amount \(\Delta F\), the frequency \(f\) changes by a small amount \(\Delta f\). Show that $$ \frac{\Delta f}{f}=\frac{1}{2} \frac{\Delta f}{f} $$ (c) The string is tuned as in part (a) when its temperature is \(18.5^{\circ} \mathrm{C}\). Strenuous playing can make the temperature of the string rise, changing its vibration frequency. Find \(\Delta f\) if the temperature of the string rises to \(29.5^{\circ} \mathrm{C}\). The steel string has a Young's modulus of \(2.00 \times 10^{11} \mathrm{~Pa}\) and a coefficient of linear expansion of \(1.20 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1} .\) Assume that the temperature of the body of the guitar remains constant. Will the vibration frequency rise or fall?

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