Problem 48 - University Physics with Modern Physics

Question

A transverse wave on a rope is given by $$y(x, t) = (0.750 \, \mathrm{cm}) \mathrm{cos} \space \pi[(10.400 \, \mathrm{cm}^{-1})x + (250 \mathrm s^{-1})t]$$ (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at these values of \(t:\) 0, 0.0005 s, 0.0010 s. (c) Is the wave traveling in the \(+x-\) or \(-x\)-direction? (d) The mass per unit length of the rope is 0.0500 kg/m. Find the tension. (e) Find the average power of this wave.

Step-by-Step Solution

Answer

(a) A = 0.750 cm, T ≈ 0.0251 s, f ≈ 39.8 Hz, λ ≈ 0.604 cm, v ≈ 24.027 cm/s. (b) See sketched shapes. (c) -x direction. (d) T ≈ 2.89 N. (e) P ≈ 4.23 W.

Step 1: Identify Wave Parameters from Equation

The given wave equation is written as: \[ y(x, t) = (0.750 \, \mathrm{cm}) \cos \left(\pi\left[(10.400 \, \mathrm{cm}^{-1})x + (250 \, \mathrm{s}^{-1})t \right]\right) \] Comparing it with the standard wave form \( y(x,t) = A \cos(kx + \omega t) \), we identify the amplitude \(A = 0.750 \, \mathrm{cm} \), the wave number \(k = 10.400 \, \mathrm{cm}^{-1} \), and the angular frequency \(\omega = 250 \, \mathrm{s}^{-1} \).

Step 2: Calculate the Amplitude

The amplitude \( A \) is the coefficient part of the cosine in the equation. Thus, \[ A = 0.750 \, \mathrm{cm} \].

Step 3: Calculate the Wavelength

The wave number \( k \) is related to the wavelength \( \lambda \) by the formula \( k = \frac{2\pi}{\lambda} \). Solving for \( \lambda \) gives:\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{10.400} \approx 0.604 \, \mathrm{cm} \].

Step 4: Calculate the Period

The angular frequency \( \omega \) is related to the period \( T \) by \( \omega = \frac{2\pi}{T} \). Solving for \( T \) gives:\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{250} \approx 0.0251 \, \mathrm{s} \].

Step 5: Calculate the Frequency

Frequency \( f \) is the reciprocal of the period. Hence,\[ f = \frac{1}{T} = \frac{1}{0.0251} \approx 39.8 \, \mathrm{Hz} \].

Step 6: Calculate Wave Speed

The wave speed \( v \) is the product of wavelength \( \lambda \) and frequency \( f \). Thus,\[ v = \lambda f = 0.604 \, \mathrm{cm} \times 39.8 \, \mathrm{Hz} \approx 24.027 \, \mathrm{cm/s} \].

Step 7: Determine Wave Direction

The term \([kx + \omega t]\) in the cosine argument indicates the wave is traveling in the \(-x\)-direction because the sign is positive (in the form of \(kx - \omega t\), the negative would indicate \(+x\)-direction).

Step 8: Sketch the Shape of the Rope

For times \( t = 0, 0.0005, \text{and} \ 0.0010 \, \mathrm{s} \), calculate \( y(x, t) \) at these times and sketch the wave shape, considering shifts due to time differences.

Step 9: Find Tension in the Rope

The wave speed \( v \) is related to tension \( T \) and linear mass density \( \mu \) by \( v = \sqrt{\frac{T}{\mu}} \). Solving for \( T \) by rearranging gives:\[ T = \mu v^2 = 0.0500 \, \mathrm{kg/m} \times (0.24027 \, \mathrm{m/s})^2 \approx 2.89 \, \mathrm{N} \].

Step 10: Calculate Average Power of the Wave

The average power \( P \) is given by \( P = \frac{1}{2} \mu v \omega^2 A^2 \). Substituting the known values gives:\[ P = \frac{1}{2} \times 0.0500 \, \mathrm{kg/m} \times 0.24027 \, \mathrm{m/s} \times (250 \, \mathrm{s}^{-1})^2 \times (0.0075 \, \mathrm{m})^2 \approx 4.23 \, \mathrm{W} \].

Key Concepts

Amplitude and Frequency

Amplitude and frequency are two key characteristics of waves. The amplitude represents the maximum displacement of a wave from its rest position. In the exercise, it is given as 0.750 cm. This is easily found by inspecting the wave equation's cosine's coefficient. The amplitude is a measure of the wave's height, indicating how far the wave's peak is from its baseline.
If you look at the wave equation, it relates amplitude to the energy a wave carries. The larger the amplitude, the more energy the wave transmits.
Frequency, on the other hand, is the number of cycles a wave completes in one second. It is found by taking the reciprocal of the wave's period. For our specific problem, the period was calculated to be approximately 0.0251 seconds. Thus, the frequency is around 39.8 Hz. Frequency indicates how fast a wave oscillates and is crucial in determining the type of sound or light emitted by the wave.

Wave Speed Calculation

Understanding wave speed is crucial as it determines how fast a wave travels through a medium. The wave speed, for this problem, is calculated using the relation between frequency and wavelength. The product of these two factors gives the wave speed:
- Wavelength (\(\lambda\)) measures the distance between consecutive wave crests, calculated as approximately 0.604 cm.
- Frequency (\(f\)) is approximately 39.8 Hz, as discussed previously.
For the given wave, these yield a wave speed, \(v\), of approximately 24.027 cm/s.
Wave speed directly depends on the properties of the medium through which the wave propagates. Different media cause waves to travel at different speeds, impacted by factors such as tension and density.

Wave Direction

The direction in which a wave travels determines how it affects its surroundings. In the context of the given exercise, determining the wave direction involves examining the sign in the wave equation. The argument of the cosine function in the wave equation uses the form \(kx + \omega t\), which indicates motion in the negative x-direction.
When you observe the formulation \(kx + \omega t\), the positive sign implies that as time progresses, the wave appears to move backwards, relative to the x-axis. Understanding wave direction is pivotal for predicting how waves interact with boundaries, other waves, and objects.

Tension in a Rope

Tension in a rope is a critical factor influencing how waves, like the one described, propagate through it. Tension refers to the force exerted along the rope that keeps it taut, and can be calculated using wave speed and the mass per unit length of the rope. For example:
- The wave speed \(v\) is approximately 24.027 cm/s.
- The linear mass density \(\mu\) is given as 0.0500 kg/m.
Using the relation \(v = \sqrt{\frac{T}{\mu}}\), we solve for tension, \(T\), which approximately equals 2.89 N in this case.
Tension affects the ease with which waves can travel, with greater tension allowing for faster wave speeds. This is crucial in applications ranging from musical instruments to engineering.

Average Power of a Wave

The average power of a wave indicates the rate at which energy is transported by the wave through the medium. In the exercise, it is given by the formula:
\[ P = \frac{1}{2} \mu v \omega^2 A^2 \]
where:

• \(\mu\) is the linear mass density (0.0500 kg/m).
• \(v\) is the wave speed (0.24027 m/s).
• \(\omega\) is the angular frequency (250 s\(^{-1}\)).
• \(A\) is the amplitude (0.0075 m).

Substituting these values results in 4.23 W, highlighting the power with which this wave transmits energy. Understanding wave power is essential for harnessing waves in technologies such as acoustics, seismology, and telecommunications.