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

The given information is as follows: 1. The wave speed is 30.0 m/s. 2. The wave has a frequency of 50.0 Hz. 3. The vertical displacement at x=0 and t=0 is 4.00 mm. 4. The wave velocity at x=0 and t=0 is 2.50 m/s.

Step-by-step solution

Determine the Period

The period of the wave (T) can be found using the frequency, as the formula is \(T = \frac{1}{f}\). The frequency is given as \(50.0 \mathrm{~Hz}\), so we calculate the period: $$ T = \frac{1}{50.0 \mathrm{~Hz}} = 0.02 \mathrm{~s} $$

Determine the Wavelength

The speed (v) of a wave is the product of its frequency (f) and wavelength (λ): $$ v= f\lambda $$ Using the given speed of \(30.0 \mathrm{~m} / \mathrm{s}\) and frequency of \(50.0 \mathrm{~Hz}\), we can find the wavelength: $$ \lambda = \frac{v}{f} = \frac{30.0 \mathrm{~m} / \mathrm{s}}{50.0 \mathrm{~Hz}} = 0.6 \mathrm{~m} $$

Determine the Amplitude

At \(x=0\) and \(t=0\), the displacement of the wave is given as \(4.00 \mathrm{~mm}\). Since this is a maximum displacement, it corresponds to the amplitude, A: $$ A = 4.00 \mathrm{~mm} = 0.004 \mathrm{~m} $$

Find the Phase Shift and Write the Wave Function

The wave is traveling in the positive \(x\)-direction, so we will use the following wave function formula: $$ y(x, t) = A \sin(kx - \omega t + \phi) $$ where \(k=\frac{2\pi}{\lambda}\) is the wave number, \(\omega = 2\pi f\) is the angular frequency, and \(\phi\) is the phase shift value. At \(x=0\) and \(t=0\), we have: $$ y = A\sin(\phi) = 0.004 $$ Since the sine function ranges between -1 and 1, the phase shift value for this case is \(\phi = 0\). This simplifies our wave function to: $$ y(x, t) = A \sin(kx - \omega t) $$ Now, with all the known values of \(A\), \(k\), and \(\omega\), we can write the complete wave function: $$ y(x, t) = 0.004 \sin \left( \frac{2\pi}{0.6} x - 2\pi \cdot 50.0 t \right) $$ Thus, the wave function for this problem is: $$ y(x, t) = 0.004 \sin \left( \frac{10\pi}{3} x - 100\pi t \right) $$

Key concepts

Wave Period

The wave period, denoted as 'T', is a fundamental concept in wave physics which refers to the time it takes for one complete cycle of the wave to pass a given point. Imagine watching a single peak on a water wave as it goes by; the wave period is the time from when the peak passes until the next peak arrives. The formula to find the period of a wave is expressed as \( T = \frac{1}{f} \) where 'f' represents the wave frequency.

In the example of our exercise, the frequency of the wave is given as 50.0 Hz, which means there are 50 cycles per second. Using the formula, we calculate the wave period as \( T = \frac{1}{50.0 \, \text{Hz}} = 0.02 \, \text{s} \). This time interval is critical for understanding the temporal behavior of waves and their interactions with physical systems.

Wave Velocity

Wave velocity, often represented by 'v', is the speed at which a wave propagates in a medium. For any given wave, whether it’s a sound wave, a water wave, or a string vibration, it tells how fast the wave fronts are moving. The wave velocity can be calculated by the formula \( v = f\lambda \), where 'f' is the frequency of the wave, and '\lambda' (lambda) is the wavelength – distance between consecutive identical points of the wave pattern, such as crest to crest or trough to trough.

For the given exercise, the wave travels at a velocity of 30.0 m/s with a frequency of 50.0 Hz. This information is used to determine the wavelength and subsequently understand how far apart are the peaks of the wave.

Amplitude of a Wave

The amplitude of a wave is the maximum displacement of points on a wave from their equilibrium positions. When we think of the height of a wave, this is what we're considering. It's a measure of how 'strong' or 'intense' the wave is, with larger amplitudes corresponding to more energetic waves. The amplitude is often denoted by 'A'.

In terms of physical observations, if one looks at a string vibrating up and down, the amplitude would be the distance from the center line (equilibrium position) to the peak of the wave. The exercise provides us with a vertical displacement of 4.00 mm at maximum, indicating that the amplitude is \( A = 4.00 \, \text{mm} = 0.004 \, \text{m} \). Amplitude is an essential parameter because it’s directly related to the energy carried by the wave; the energy is proportional to the square of the wave's amplitude.

Wave Frequency

Wave frequency is defined as the number of wave cycles that pass a point in one second and is represented by the symbol 'f'. Frequency is measured in Hertz (Hz), which is equivalent to one cycle per second. Determining the wave frequency is vital as it directly correlates with the energy of the wave; higher frequency waves carry more energy.

In our exercise, the frequency has been given as 50.0 Hz, meaning 50 wave cycles occur each second. This frequency, alongside other wave properties, is used to derive the wave function that mathematically describes the wave's behavior over time and position. A higher frequency means that the wave oscillates more quickly, and this concept is essential when exploring the nature of waves and their interactions with the environment.