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Chapter 17: Problem 3

Sinusoidal Wave A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is \(0.170 \mathrm{~s}\). What are the (a) period and (b) frequency? (c) The wave length is \(1.40 \mathrm{~m}\); what is the wave speed?

Short Answer

Expert verified
a) 0.680 s, b) 1.47 Hz, c) 2.06 m/s

Step by step solution

01

Understand the given information

The time for a point to move from maximum displacement to zero displacement (i.e., a quarter period) is given as 0.170 s. The wavelength \(\lambda\) is given as 1.40 m.
02

Calculate the period

The time given is one-fourth of the period (T). Therefore, the full period \ (T) \ is calculated as \[ T = 4 \times 0.170 \mathrm{s} = 0.680 \mathrm{s} \].
03

Calculate the frequency

Frequency \( f \) is the reciprocal of the period \( T \). Thus, \[ f = \frac{1}{T} = \frac{1}{0.680 \mathrm{s}} = 1.47 \mathrm{Hz} \].
04

Calculate the wave speed

Wave speed \( v \) is given by \ v = f \lambda \. Substituting the known values, \[ v = 1.47 \mathrm{Hz} \times 1.40 \mathrm{m} = 2.06 \mathrm{m/s} \].

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Key Concepts

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

Period of a Wave
The period of a wave, often represented as T, is the time it takes for a single cycle of the wave to complete. In other words, it's the time for one full oscillation. Understanding the period is crucial because it helps us understand how frequently the wave cycles over time. In the given exercise, we know that the time for the wave to move from its maximum displacement to zero (which is a quarter of a full cycle) is 0.170 seconds. To find the full period, we simply multiply this value by 4, giving us \[ T = 4 \times 0.170 \text{ s} = 0.680 \text{ s} \]. This tells us that one complete cycle of the wave takes 0.680 seconds. By understanding the period, we can also derive other critical properties of the wave, such as its frequency.
Frequency Calculation
Frequency refers to how many cycles of a wave pass a particular point in one second, and it鈥檚 measured in Hertz (Hz). It is the reciprocal of the period, which means if you know the period (T), you can easily find the frequency (f) using the formula \[ f = \frac{1}{T} \]. In our example, the period of the wave is 0.680 seconds. Using this, we calculate the frequency as \[ f = \frac{1}{0.680 \text{ s}} = 1.47 \text{ Hz} \]. This means that in one second, 1.47 cycles of the wave pass by a given point. Understanding frequency is essential in areas such as signal processing, sound engineering, and many other fields.
Wave Speed Determination
The speed of a wave (v) tells us how fast the wave travels through a medium. To find the wave speed, we use the relationship between frequency (f) and wavelength (\( \lambda \)). The formula for wave speed is \[ v = f \lambda \]. From the exercise, we know that the frequency is 1.47 Hz and the wavelength is 1.40 meters. Substituting these values into our formula, we get \[ v = 1.47 \text{ Hz} \times 1.40 \text{ m} = 2.06 \text{ m/s} \]. This result means that the wave travels at a speed of 2.06 meters per second through the medium. Calculating the wave speed is particularly important in understanding how different media affect wave propagation, such as how sound waves travel faster in water than in air.

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Most popular questions from this chapter

Equation of a Transverse The equation of a transverse wave traveling along a very long string is \(y(x, t)=(6.0 \mathrm{~cm}) \sin \\{(0.020 \mathrm{\pi} \mathrm{rad} / \mathrm{cm})\) \(x+(4.0 \pi \mathrm{rad} / \mathrm{s}) t]\) where \(x\) and \(y\) are expressed in centimeters and \(t\) is in seconds Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at \(x=3.5 \mathrm{~cm}\) when \(t=0.26 \mathrm{~s}\) ?

What Is the Speed What is the speed of a transverse wave in a rope of length \(2.00 \mathrm{~m}\) and mass \(60.0 \mathrm{~g}\) under a tension of \(500 \mathrm{~N}\) ?

Sinusoidal Wave Two A sinusoidal wave of frequency \(500 \mathrm{~Hz}\) has a speed of \(350 \mathrm{~m} / \mathrm{s}\). (a) How far apart are two points that differ in phase by \(\pi / 3 \mathrm{rad} ?(\mathrm{~b})\) What is the phase difference between two displacements at a certain point at times \(1.00 \mathrm{~ms}\) apart?

Three Lowest Frequencies What are the three lowest frequencies for standing waves on a wire \(10.0 \mathrm{~m}\) long having a mass of \(100 \mathrm{~g}\), which is stretched under a tension of \(250 \mathrm{~N} ?\)

Linear Density The linear density of a string is \(1.6 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). A transverse wave on the string is described by the equation $$ y(x, t)=(0.021 \mathrm{~m}) \sin [(2.0 \mathrm{rad} / \mathrm{m}) x+(30 \mathrm{rad} / \mathrm{s}) t] $$ What is (a) the wave speed and (b) the tension in the string?
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