Problem 62
Question
A sinusoidal transverse wave traveling in the positive direction of an \(x\) axis has an amplitude of \(2.0 \mathrm{~cm}\), a wavelength of \(10 \mathrm{~cm}\), and a frequency of \(400 \mathrm{~Hz}\). If the wave equation is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (a) \(y_{m}\), (b) \(k\), (c) \(\omega\), and (d) the correct choice of sign in front of \(\omega ?\) What are (e) the maximum transverse speed of a point on the cord and (f) the speed of the wave?
Step-by-Step Solution
Verified Answer
a) 2.0 cm, b) \(\pi/5 \ cm^{-1}\), c) \(800\pi \ rad/s\), d) negative sign, e) \(16\pi \ cm/s\), f) 4000 cm/s.
1Step 1: Identifying Amplitude
The amplitude of the wave, denoted by \( y_m \), is the maximum vertical displacement of the wave from its resting position. It is given as \( 2.0 \ cm \). Thus, \( y_m = 2.0 \ cm \).
2Step 2: Calculating Wavenumber
The wavenumber \( k \) describes how many cycles fit into a unit length, defined as \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength. The wavelength given is \( 10 \ cm \). Therefore, \( k = \frac{2\pi}{10 \ cm} = \frac{\pi}{5} \ cm^{-1} \).
3Step 3: Finding Angular Frequency
The angular frequency \( \omega \) is related to the frequency \( f \) by the equation \( \omega = 2\pi f \). With the frequency given as \( 400 \ Hz \), it follows that \( \omega = 2\pi \times 400 = 800\pi \ rad/s \).
4Step 4: Choosing the Correct Sign for Wave Equation
Since the wave is traveling in the positive direction of the \( x \) axis, the wave equation should be in the form \( y(x, t) = y_m \sin(kx - \omega t) \) with the negative sign.
5Step 5: Calculating Maximum Transverse Speed
The maximum transverse speed of a point on the cord is \( v_{\text{max}} = \omega y_m \). Substituting the values, \( v_{\text{max}} = 800\pi \times 2.0 \times 10^{-2} = 16\pi \ cm/s \approx 50.27 \ cm/s \).
6Step 6: Determining Wave Speed
The speed of the wave \( v \) can be determined by the relationship \( v = f\lambda \). Substituting the given values, \( v = 400 \times 10 = 4000 \ cm/s \).
Key Concepts
AmplitudeWavelengthFrequencyAngular Frequency
Amplitude
In the study of waves, the amplitude is a fundamental concept that reflects the wave's intensity or strength. The amplitude is the maximum extent of a wave measured from its equilibrium position. In mathematical terms, it's the height of the wave peaks or the depth of the troughs from the rest position. For a sinusoidal wave such as the one described in the exercise, the amplitude ( \( y_m \)) is strictly the maximum displacement of particles from the wave's rest position.
In this exercise, the amplitude is given as \(2.0 \, \mathrm{cm}\). It signifies how far the particles are moving from their equilibrium, suggesting that every crest and trough is \(2.0 \, \mathrm{cm}\) away from the rest line. This concept is critical in different realms such as physics and engineering because:
In this exercise, the amplitude is given as \(2.0 \, \mathrm{cm}\). It signifies how far the particles are moving from their equilibrium, suggesting that every crest and trough is \(2.0 \, \mathrm{cm}\) away from the rest line. This concept is critical in different realms such as physics and engineering because:
- It determines the wave's energy, as energy is proportional to the square of the amplitude.
- It affects how we perceive sound's loudness or light's brightness, where higher amplitudes mean louder sounds or brighter lights.
Wavelength
The wavelength of a wave is the distance over which the wave's shape repeats, essentially measuring the length of one complete cycle of the wave. Wavelength is usually denoted by the Greek letter (\( \lambda \)) and in our exercise, it is given as \(10 \, \mathrm{cm}\).
Understanding wavelength is crucial because:
Understanding wavelength is crucial because:
- It plays a role in determining the wave's wave number (\(k\)) using the formula \(k = \frac{2\pi}{\lambda}\).
- It influences the wave's speed (\(v\)), which can be calculated by multiplying wavelength with frequency, \(v = f \lambda\).
- It is integral to the properties of various types of waves, including electromagnetic and sound waves, influencing how they behave and interact with materials.
Frequency
Frequency is all about how often something happens over a specified time period. In the context of wave motion, it is the number of wave cycles that pass a point per unit time, usually one second, and is denoted by the symbol \(f\). The unit of frequency is hertz (\(Hz\)), where 1 Hz equals one cycle per second.
In the exercise example, a frequency of \(400 \, \text{Hz}\) is provided, indicating that \(400\) cycles pass a point each second. The concept of frequency is vital because:
In the exercise example, a frequency of \(400 \, \text{Hz}\) is provided, indicating that \(400\) cycles pass a point each second. The concept of frequency is vital because:
- It relates to angular frequency (\(\omega\)) through \(\omega = 2\pi f\).
- It influences the pitch of sound in acoustics—a higher frequency means a higher pitch.
- In electronics and signal processing, frequency is key to understanding alternating current (AC) signals and oscillations.
Angular Frequency
Angular frequency is a concept that applies to oscillatory motion, representing how fast something rotates or oscillates in terms of angles, typically in radians per second. It’s closely related to the basic concept of frequency by the relation \(\omega = 2\pi f\), where \(\omega\) represents angular frequency. In our exercise, with a given frequency of \(400 \, \text{Hz}\), the angular frequency becomes \(800\pi \, \text{rad/s}\).
Angular frequency is essential to understand because:
Angular frequency is essential to understand because:
- It transforms the cyclic frequency (in hertz) to angular terms (in radians), crucial for analyzing rotational motions and oscillations in physics.
- It plays a significant part in the wave equation \(y(x, t) = y_m \sin(kx - \omega t)\), influencing both calculations of speed (\(v_{\text{max}} = \omega y_m\)) and phase shifts.
- Systems described by sinusoidal functions use angular frequency to express time variations compactly.
Other exercises in this chapter
Problem 57
A generator at one end of a very long string creates a wave given by $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x+\left(
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In Fig. \(16-42\), a string, tied to a sinusoidal oscillator at \(P\) and running over a support at \(Q\), is stretched by a block of mass \(m\). The separation
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A wave has a speed of \(240 \mathrm{~m} / \mathrm{s}\) and a wavelength of \(3.2 \mathrm{~m}\). What are the (a) frequency and (b) period of the wave?
View solution Problem 64
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
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