Question

A wave on a string has a wave function given by $$ y(x, t)=(0.0200 \mathrm{~m}) \sin \left[\left(6.35 \mathrm{~m}^{-1}\right) x+\left(2.63 \mathrm{~s}^{-1}\right) t\right] $$ a) What is the amplitude of the wave? b) What is the period of the wave? c) What is the wavelength of the wave? d) What is the speed of the wave? e) In which direction does the wave travel?

Short answer

Question: Determine the properties of the wave given by the wave equation as follows: $$ y(x, t) = (0.0200~m) \sin \left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right] $$ Properties to find: a) Amplitude b) Period c) Wavelength d) Speed of the wave e) Direction of the wave Answer: a) Amplitude: \(A = 0.0200~m\) b) Period: \(T \approx 2.39~s\) c) Wavelength: \(\lambda \approx 0.990~m\) d) Speed of wave: \(v \approx 0.414~m/s\) e) Direction of wave: negative x-direction

Step-by-step solution

Identify the amplitude

From the given wave equation, $$ y(x, t) = (0.0200~m) \sin \left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right] $$ Compare this equation with the general wave equation, we can immediately determine the amplitude as: $$ A = 0.0200~m $$

Identify the angular frequency and period

From the given wave equation, $$ y(x, t) = (0.0200~m) \sin \left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right] $$ We can determine the angular frequency (ω) as: $$ \omega = 2.63~s^{-1} $$ To calculate the period (T) of the wave, we can use the following relation between angular frequency and period: $$ \omega = \frac{2\pi}{T} $$ Now solve for the period T: $$ T = \frac{2\pi}{\omega} = \frac{2\pi}{2.63~s^{-1}} \approx 2.39~s $$

Identify the wave number and wavelength

From the given wave equation, $$ y(x, t) = (0.0200~m) \sin \left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right] $$ We can determine the wave number (k) as: $$ k = 6.35~m^{-1} $$ To calculate the wavelength (λ) of the wave, we can use the following relation between wave number and wavelength: $$ k = \frac{2\pi}{\lambda} $$ Now solve for the wavelength λ: $$ \lambda = \frac{2\pi}{k} = \frac{2\pi}{6.35~m^{-1}} \approx 0.990~m $$

Calculate the speed of the wave

The speed of the wave (v) can be calculated using the following relation between wavelength and period: $$ v = \frac{\lambda}{T} $$ Now substitute the previously calculated values of λ and T, $$ v = \frac{0.990~m}{2.39~s} \approx 0.414~m/s $$

Determine the direction of the wave

From the given wave equation, $$ y(x, t) = (0.0200~m) \sin \left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right] $$ The wave function has a positive sign between the \(kx\) and \(ωt\) terms, indicating a wave traveling in the negative x-direction. The general form with a positive sign is: $$ y(x,t) = A \sin(kx - \omega t) $$ So, the wave travels in the negative x-direction.

Summary

Here are the answers to the problem: a) Amplitude: \(A = 0.0200~m\) b) Period: \(T \approx 2.39~s\) c) Wavelength: \(\lambda \approx 0.990~m\) d) Speed of wave: \(v \approx 0.414~m/s\) e) Direction of wave: negative x-direction

Key concepts

Wave Amplitude

The wave amplitude is a measure of the height of the wave. It's like how tall a wave is from its resting position. In mathematical terms, it's the maximum displacement of the wave from its equilibrium position. When looking at the wave function given in the problem,

• \(y(x, t) = (0.0200~m) \sin\left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right]\)

You can see that the wave amplitude is represented by the constant in front of the sine function. Here, the amplitude is

• \(A = 0.0200~m\)

This tells us that the wave reaches up to 0.0200 meters above and below its central rest position. Essentially, if you imagine the wave on a graph, it would move 0.0200 meters up and 0.0200 meters down from the middle line.

Wave Period

The wave period is how long it takes for one complete cycle of the wave to pass a given point. It measures time, typically in seconds, for a repeating pattern to completely repeat itself.
In the wave function

• \(y(x, t) = (0.0200~m) \sin\left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right]\)

You can determine the period from the angular frequency \(\omega\). Here, \(\omega = 2.63~s^{-1}\). To find the period \(T\), use the formula

• \(\omega = \frac{2\pi}{T}\)

Solving for \(T\), we find

• \(T = \frac{2\pi}{2.63~s^{-1}} \approx 2.39~s\)

This means each wave cycle takes approximately 2.39 seconds to complete.

Wave Wavelength

The wave wavelength is the distance between two corresponding points on consecutive cycles of the wave, such as peak to peak or trough to trough. It's essentially the length of one complete wave cycle.
Given the wave function

• \(y(x, t) = (0.0200~m) \sin\left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right]\)

The wave number \(k\) provides us with this information. Here, \(k = 6.35~m^{-1}\). To find the wavelength \(\lambda\), use the relationship

• \(k = \frac{2\pi}{\lambda}\)

Solving for \(\lambda\), we find

• \(\lambda = \frac{2\pi}{6.35~m^{-1}} \approx 0.990~m\)

Thus, the wavelength is approximately 0.990 meters, indicating how long one wave cycle is in physical space.

Wave Speed

Wave speed tells us how fast the wave is moving, which in turn shows how quickly energy is transferred by the wave. It's calculated by multiplying the frequency and wavelength.
In our example, having calculated

• Wavelength \(\lambda = 0.990~m\)
• Period \(T \approx 2.39~s\)

The wave speed \(v\) can be found using the formula

• \(v = \frac{\lambda}{T}\)

Substitute the values to find

• \(v = \frac{0.990~m}{2.39~s} \approx 0.414~m/s\)

This means the wave travels at a speed of approximately 0.414 meters per second.

Wave Direction

Wave direction indicates the path along which the wave energy moves. It is crucial as it impacts how energy is distributed in the medium through which the wave is moving.
The sign in front of the term combinations \(kx\) and \(\omega t\) in a wave function often hints at direction. For the given wave equation

• \(y(x, t) = (0.0200~m) \sin\left[ (6.35~m^{-1}) x + (2.63~s^{-1}) t \right]\)

The positive sign between \(kx\) and \(\omega t\) points to the wave moving in the negative x-direction. Typically, the familiar form of a rightward wave is \(y(x,t) = A \sin(kx - \omega t)\).
Since our function instead resembles \(y(x,t) = A \sin(kx + \omega t)\), this indicates movement in the opposite direction, confirming the wave is traveling towards the negative x-direction.