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Chapter 16: Problem 40

Two sinusoidal waves with identical wavelengths and amplitudes travel in opposite directions along a string with a speed of \(10 \mathrm{~cm} / \mathrm{s}\). If the time interval between instants when the string is flat is \(0.50 \mathrm{~s}\), what is the wavelength of the waves?

Short Answer

Expert verified
The wavelength of the waves is 10 cm.

Step by step solution

01

Understanding Wave Properties

Two sinusoidal waves are traveling in opposite directions, resulting in a standing wave. The speed of these waves is given as \(10 \text{ cm/s}\). Further, the time interval between each instance of the string being flat (node or zero displacement) is provided as \(0.50 \text{ s}\). We need to find the wavelength of these waves.
02

Identifying Time Between Consecutive Zero Displacements

The string is flat at nodes. The given time interval of \(0.50 \text{ s}\) corresponds to half the period of the wave, since it is the time between consecutive zero displacements.
03

Calculate the Wave Period

Since the time interval of \(0.50 \text{ s}\) represents half of the wave's period, the full period \(T\) is double this value:\[ T = 2 imes 0.50 \text{ s} = 1.00 \text{ s}. \]
04

Relate Wave Speed and Wavelength

The wave speed \(v\), period \(T\), and wavelength \(\lambda\) are related by the equation: \[ v = \frac{\lambda}{T}. \]We know \(v = 10 \text{ cm/s}\) and \(T = 1.00 \text{ s}\).
05

Solve for Wavelength

Rearrange the equation \( v = \frac{\lambda}{T} \) to solve for the wavelength:\[ \lambda = v \cdot T. \]Substitute in the given values:\[ \lambda = 10 \text{ cm/s} \cdot 1.00 \text{ s} = 10 \text{ cm}. \]

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Key Concepts

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

Wave Properties
When looking at wave properties, it's important to know what these key characteristics tell us about the wave. For sinusoidal waves, like the ones in this exercise, properties include amplitude, wavelength, frequency, speed, and often phase. In our scenario, we have two waves moving in opposing directions along a string. They interfere to create a standing wave. This type of wave appears to be stationary, although in reality, it's made up of two superimposed traveling waves. The fixed points where the wave appears untouched, or 'flat,' are called nodes.

Two essential properties are highlighted here: the wave speed (10 cm/s) and the time between zero displacements (0.50 s). Understanding these properties allows us to explore deeper into the behavior of these waves.
Wavelength Calculation
Calculating the wavelength (\(\lambda\)) can be easily done when you know the wave speed and period. This exercise approached it by first identifying the importance of the time interval of 0.50 s鈥攖his is half of the wave's full period because it's the gap between nodes. Recognizing this helps to find the entire period.To find the wavelength, use the formula:
  • \(\lambda = v \cdot T\)
where\(v\) is the wave speed and\(T\) is the period. Once \(T\) is calculated, simply multiply by the wave speed to get the wavelength鈥攁s worked out in this case to be 10 cm.
Wave Speed
Wave speed (\(v\)) represents how fast the wave is traveling through the medium鈥攊n this case, a string. For sinusoidal waves, wave speed depends on both the medium and wave properties. Here, it is given as 10 cm/s, making calculations especially straightforward.Wave speed can be defined through the relationship:
  • \(v = \frac{\lambda}{T}\)
This equation comes from dividing the wavelength by the period. It illustrates that if you know any two of these properties, you can find the third. Recognizing these relationships can improve your intuition about wave behavior and aid in solving exercises efficiently.
Wave Period
The wave period (\(T\)) is the time taken for one complete cycle of the wave. It鈥檚 a key concept to grasp, especially in exercises involving periodicity, like this one. Knowing the period can help in calculating other properties such as frequency and wavelength.In this scenario, the provided time interval (0.50 s) was particularly useful. It represents half of the wave's period because it denotes the time between successive moments of zero displacement. Therefore, to find the complete period, multiply the given time by two:
  • \(T = 2 \times 0.50 \, \text{s} = 1.00 \, \text{s}\)
This insight was critical for calculating the wavelength and understanding the overall behavior of the standing wave on the string.

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Most popular questions from this chapter

The equation of a transverse wave traveling along a string is $$ y=0.15 \sin (0.79 x-13 t) $$ in which \(x\) and \(y\) are in meters and \(t\) is in seconds. (a) What is the displacement \(y\) at \(x=2.3 \mathrm{~m}, t=0.16 \mathrm{~s} ?\) A second wave is to be added to the first wave to produce standing waves on the string. If the second wave is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (b) \(y_{m},(\mathrm{c})\) \(k\), (d) \(\omega\), and (e) the correct choice of sign in front of \(\omega\) for this second wave? (f) What is the displacement of the resultant standing wave at \(x=2.3 \mathrm{~m}, t=0.16 \mathrm{~s}\) ?

A uniform rope of mass \(m\) and length \(L\) hangs from a ceiling. (a) Show that the speed of a transverse wave on the rope is a function of \(y\), the distance from the lower end, and is given by \(v=\sqrt{g y}\). (b) Show that the time a transverse wave takes to travel the length of the rope is given by \(t=2 \sqrt{L / g} .\)

The type of rubber band used inside some baseballs and golf balls obeys Hooke's law over a wide range of elongation of the band. A segment of this material has an unstretched length \(\ell\) and a mass \(m .\) When a force \(F\) is applied, the band stretches an additional length \(\Delta \ell .\) (a) What is the speed (in terms of \(m, \Delta \ell\), and the spring constant \(k\) ) of transverse waves on this stretched rubber band? (b) Using your answer to (a), show that the time required for a transverse pulse to travel the length of the rubber band is proportional to \(1 / \sqrt{\Delta \ell}\) if \(\Delta \ell \& \ell\) and is constant if \(\Delta \ell\) ? \(\ell\).

A string that is stretched between fixed supports separated by \(75.0 \mathrm{~cm}\) has resonant frequencies of 420 and \(315 \mathrm{~Hz}\), with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Two identical traveling waves, moving in the same direction, are out of phase by \(\pi / 2\) rad. What is the amplitude of the resultant wave in terms of the common amplitude \(y_{m}\) of the two combining waves?
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