Question

A sinusoidal wave on a string is described by the equation \(y=(0.100 \mathrm{~m}) \sin (0.750 x-40.0 t),\) where \(x\) and \(y\) are in meters and \(t\) is in seconds. If the linear mass density of the string is \(10.0 \mathrm{~g} / \mathrm{m},\) determine (a) the phase constant, (b) the phase of the wave at \(x=2.00 \mathrm{~cm}\) and \(t=0.100 \mathrm{~s}\), (c) the speed of the wave, (d) the wavelength, (e) the frequency, and (f) the power transmitted by the wave.

Short answer

Answer: The phase of the wave at position x = 2.00 cm and time t = 0.100 s is -4.00.

Step-by-step solution

Identify and calculate the phase constant

The phase constant is the constant term added inside the sinusoidal function. In our case, the wave equation is given by: \(y = (0.100 \mathrm{~m}) \sin(0.750x - 40.0t)\) The phase constant is the value -40.0t, so we can write that \(\phi = -40.0t\).

Calculate the phase of the wave at a specific position and time

To calculate the phase at position x = 2.00 cm and time t = 0.100 s, we plug in these values into the phase term we found in step 1: \(\phi(x,t) = -40.0 (0.100 \mathrm{~s}) = -4.00\)

Determine the wave's speed

The speed of the wave can be found from the coefficient of the time term in the phase. We have: \(\omega = 40.0 \mathrm{s}^{-1}\) The wave speed \(v\) is related to \(\omega\) and \(k\) (the coefficient of the x term) by \(v = \frac{\omega}{k}\). So, we can write: \(v = \frac{40.0 \mathrm{s}^{-1}}{0.750 \mathrm{m}^{-1}} = 53.3 \mathrm{m/s}\)

Calculate the wavelength

The wavelength \(\lambda\) is related to the wave number \(k\) by \(\lambda = \frac{2\pi}{k}\). So, we can write: \(\lambda = \frac{2\pi}{0.750 \mathrm{m}^{-1}} = 8.38 \mathrm{m}\)

Determine the frequency

The frequency \(f\) is related to the angular frequency \(\omega\) by \(f = \frac{\omega}{2\pi}\). So, we can write: \(f = \frac{40.0 \mathrm{s}^{-1}}{2\pi} = 6.37 \mathrm{Hz}\)

Calculate the power transmitted by the wave

The power transmitted by the wave can be calculated using the following formula: \(P = \frac{1}{2} \mu v^2 A^2 \omega^2\) where \(\mu\) is the linear mass density, \(A\) is the wave amplitude, and \(\omega\) is the angular frequency. We are given: \(\mu = 10.0 \mathrm{g/m} = 0.0100 \mathrm{kg/m}\) \(A = 0.100 \mathrm{m}\) \(\omega = 40.0 \mathrm{s}^{-1}\) We already found the wave speed \(v = 53.3 \mathrm{m/s}\). Now, we plug in these values into the power formula: \(P = \frac{1}{2} (0.0100 \mathrm{kg/m})(53.3 \mathrm{m/s})^2 (0.100 \mathrm{m})^2 (40.0 \mathrm{s}^{-1})^2\) \(P = 71.6 \mathrm{W}\) So, the power transmitted by the wave is 71.6 W.

Key concepts

Phase Constant

The phase constant in a sinusoidal wave equation plays a crucial role in determining the initial angle of the wave at the start of the observation. It is part of the equation that affects the horizontal shift of the wave pattern. In our exercise, the sinusoidal wave equation on a string is expressed as \(y=(0.100 \text{ m}) \sin (0.750x-40.0t)\), where \(\phi = -40.0t\) without any added constant term. This means that our phase constant \(\phi_0\) is effectively 0, as there is no separate constant value added or subtracted from the wave's expression. Therefore, the wave starts at its equilibrium position at \(t=0\).

Understanding the phase constant helps students grasp how the wave's position is affected at different times, and it is essential when studying wave interference, where the phase difference between waves can drastically change the resulting wave pattern.

Wave Speed

Wave speed is a fundamental property of a wave that describes how fast a wave travels through a medium. It is calculated from the equation of the wave, specifically from the relationship between the angular frequency \(\omega\) and the wave number \(k\). For the exercise, the angular frequency is \(\omega = 40.0 \text{s}^{-1}\) and \(k=0.750 \text{m}^{-1}\). Using the formula \(v = \frac{\omega}{k}\), we computed the wave speed to be \(v = 53.3 \text{m/s}\).

Wave speed is vital for understanding how wave properties such as wavelength and frequency are related—since \(v=\lambda f\), where \(\lambda\) is the wavelength and \(f\) is the frequency. This is why, in various applications such as communications and physics, knowing the wave speed enables us to calculate distances and signal timings effectively.

Wavelength

The wavelength, denoted by \(\lambda\), signifies the distance over which the wave's shape repeats. It is the distance between consecutive points that are in phase—meaning they are at the same stage in their oscillation. In our given problem, we determined the wavelength via the formula \(\lambda = \frac{2\pi}{k}\), which yielded a wavelength of \(\lambda = 8.38 \text{m}\).

The wavelength is directly related to the frequency of the wave, which is why these two properties are often discussed together. A shorter wavelength corresponds to a higher frequency, meaning the waves are more tightly packed, and vice versa. In practical scenarios, like sound waves or electromagnetic waves, the wavelength can determine properties like pitch and color respectively.

Frequency

Frequency is defined as the number of complete oscillations or cycles a wave undergoes per unit time, usually measured in hertz (Hz). It is calculated from the angular frequency \(\omega\) by the equation \(f = \frac{\omega}{2\pi}\). In our exercise, with an angular frequency of \(\omega = 40.0 \text{s}^{-1}\), we found the frequency of the wave to be \(f = 6.37 \text{Hz}\).

Frequency gives us an insight into how often the wave oscillates, which is essential for understanding phenomena like resonance—the tendency of a system to oscillate more powerfully at some frequencies—and wave energy. For instance, in acoustics, a higher frequency is perceived as a higher pitch.

Power Transmitted by a Wave

Power in the context of waves refers to the amount of energy that a wave carries per unit time, measured in watts (W). The power transmitted by a wave can be described using the formula \(P = \frac{1}{2} \mu v^2 A^2 \omega^2\), where \(\mu\) is the linear mass density, \(v\) is the wave speed, \(A\) is the amplitude, and \(\omega\) is the angular frequency. By substituting the given values into this formula, the exercise showed us that the power transmitted by the wave is 71.6 W.

The concept of power is crucial in many fields, from engineering and design of structures that must withstand different wave energies to the generation and transmission of electromagnetic waves in wireless communications. Understanding this aspect of wave physics can lead to the efficient transmission of energy and minimizing losses in various systems.