Question

A wave travels along a string in the positive $x$ -direction at $30.0\mathrm{m}/\mathrm{s}$. The frequency of the wave is $50.0\mathrm{Hz}$. At $x=0$ and $t=0,$ the wave velocity is $2.50\mathrm{m}/\mathrm{s}$ and the vertical displacement is $y=4.00\mathrm{mm}.$ Write the function $y(x,t)$ for the wave

Short answer

In this exercise, we found the function y(x, t) that describes a wave traveling along a string in the positive x-direction. We were given the wave velocity (30.0 m/s), frequency (50.0 Hz), and initial conditions for wave velocity and vertical displacement. We used the general equation of a wave, y(x, t) = A sin(kx - ωt + φ), and the given information to find the amplitude (A = 4.00 mm), wave number (k = (10π/3) m⁻¹), angular frequency (ω = 100π rad/s), and phase constant (φ) for this particular wave. The final y(x, t) function for this wave is: y(x, t) = 4.00 mm sin( (10π/3) x - 100πt + φ)

Step-by-step solution

Review the general equation of a wave

The general equation for a wave moving in the positive x-direction is given by: $y(x,t)=A\sin (kx-\omega t+\varphi )$ Here, $y(x,t)$: The displacement of the string at position x and time t, $A$: Amplitude of the wave, $k$: Wave number, $x$: Position on the string, $\omega$: Angular frequency, $t$: Time, $\varphi$: Phase constant.

Determine the wavelength and wave number

We are given the wave velocity ($v$) and frequency ($f$) of the wave. To calculate the wavelength ($\lambda$), we can use the formula: $v=f\lambda$ Rearranging for $\lambda$: $\lambda =\frac{v}{f}=\frac{30.0\text{m/s}}{50.0\text{Hz}}=0.6\text{m}$ Now, we can find the wave number ($k$) using the following formula: $k=\frac{2\pi }{\lambda }=\frac{2\pi }{0.6\text{m}}=\frac{10\pi }{3}{\text{m}}^{-1}$

Determine the angular frequency

We are given the frequency ($f$) of the wave and can use this information to calculate the angular frequency ($\omega$): $\omega =2\pi f=2\pi (50.0\text{Hz})=100\pi \text{rad/s}$

Determine the amplitude, phase constant, and initial conditions

We are given the initial values of displacement ($y$) and wave velocity ($u$) when $x=0$ and $t=0$. We use these values to find the amplitude ($A$) and phase constant ($\varphi$). At $x=0$ and $t=0$, $y(0,0)=A\sin (\varphi )=4.00\text{mm}$ $u(0,0)=-A\omega \cos (\varphi )=2.50\text{m/s}$ First, let's find $A$: $A=\sqrt{y^2+{\left(\frac{u}{\omega }\right)}^2}=\sqrt{(4.00\text{mm}{)}^2+{\left(\frac{2.50\text{m/s}}{100\pi \text{rad/s}}\right)}^2}$ $A=0.004\text{m}=4.00\text{mm}$ Now let's find the phase constant ($\varphi$): $\varphi =\arctan \left(\frac{y(0,0)}{u(0,0)/\omega }\right)=\arctan \left(\frac{4.00\text{mm}}{2.50\text{m/s}/100\pi \text{rad/s}}\right)$

Write the function y(x, t) for the wave

Now that we have found all the necessary values, we can substitute them back into the general equation of the wave to obtain the y(x, t) function for this specific wave: $y(x,t)=A\sin (kx-\omega t+\varphi )=4.00\text{mm}\sin (\frac{10\pi }{3}x-100\pi t+\varphi )$ Here, $\varphi$ is the angle found in Step 4.

Key concepts

Wave Velocity

Wave velocity, also referred to as phase velocity, is a measure of how fast a wave propagates through a medium. It's represented by the symbol $v$ and is typically measured in meters per second (m/s). The velocity of a wave is determined by the medium through which the wave is traveling and by the type of wave. For example, the speed of sound is different in air compared to water. In our exercise, the wave velocity is given as $30.0\text{m/s}$, indicating how fast the wave travels along the string.
Wave velocity can be mathematically related to the frequency $f$ and the wavelength $\lambda$ of the wave by the equation $v=f\lambda$. Understanding wave velocity is crucial for a variety of applications, including communication technologies where signal transmission speed is key.

Wave Frequency

The frequency of a wave, represented by $f$, is the number of oscillations or cycles that occur per unit time, and it is measured in Hertz (Hz). In simple terms, it's how many times the wave vibrates up and down in one second. Frequency is an intrinsic property of a wave and is determined by the source of the wave. In our problem, the frequency of the wave is $50.0\text{Hz}$.
A higher frequency means more waves are passing through a point each second, resulting in a higher pitch sound or, in case of light waves, a shift towards the blue end of the spectrum. On the other hand, lower frequencies correlate with lower-pitched sounds or a shift towards the red end of the light spectrum. The frequency is a fundamental property in understanding wave behaviors and in designing systems that utilize waves.

Angular Frequency

Angular frequency, denoted by $\omega$, is related to the wave frequency and represents the rate of change of the phase of a sine wave. It is connected to how rapidly the wave oscillates in radians per second and is calculated using the formula $\omega =2\pi f$, where $f$ is the frequency of the wave. In the context of our exercise, the angular frequency is $100\pi \text{rad/s}$.
The concept of angular frequency is particularly useful in physics and engineering when dealing with sinusoidal functions, oscillatory motion, and alternating currents in electrical circuits. It provides a different perspective on how we describe wave motion, emphasizing the rotational aspect of oscillations.

Wave Number

The wave number, symbolized as $k$, is a spatial analogue to the angular frequency and provides information on the wave's spatial structure. It represents the number of wave cycles per unit distance and is measured in reciprocal meters ($m^{-1}$). The mathematical relationship is given by $k=\frac{2\pi }{\lambda }$, where $\lambda$ is the wavelength of the wave. From our given problem, the wave number is calculated to be $\frac{10\pi }{3}{\text{m}}^{-1}$.
The concept of wave number is important when describing how a wave repeats itself over a certain distance and influences phenomena like diffraction and interference. Wave number connects the physical dimensions of the wave to its more commonly discussed properties, such as wavelength and frequency.

Amplitude of Wave

The amplitude of a wave, usually denoted as $A$, is the maximum displacement of points on a wave from the wave's equilibrium position. In our exercise, the amplitude of the wave is $4.00\text{mm}$. Amplitude is directly related to the energy transmitted by the wave: a higher amplitude means the wave carries more energy. For sound waves, larger amplitudes correlate with louder sounds, while for light, they correspond to greater brightness.
In wave equations, the amplitude is a measure of how far the medium (in this case, a string) is displaced in a wave pattern. It is a critical factor in various physical phenomena, from the loudness of noises to the intensity of seismic waves in earthquakes. Amplitude is a key property in understanding the energy and power of waves.