Problem 18 - University Physics with Modern Physics

Question

A 1.50 -m string of weight 0.0125 \(\mathrm{N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W\) . Neglect the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t)=(8.50 \mathrm{mm}) \cos \left(172 \mathrm{m}^{-1} x-4830 \mathrm{s}^{-1} t\right)$$ Assume that the tension of the string is constant and equal to \(W\) . (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W ?\) (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for wayes traveling down the string?

Step-by-Step Solution

Answer

(a) 0.0534 s; (b) 0.670 N; (c) 41 wavelengths; (d) Wave equation: \(y(x, t) = (8.50 \text{ mm}) \cos(-172 \text{ m}^{-1} x - 4830 \text{ s}^{-1} t)\).

Step 1: Identify Wave Variables

The given wave equation is \[y(x, t) = (8.50 \text{ mm}) \cos(172 \text{ m}^{-1} x - 4830 \text{ s}^{-1} t)\]From this, we extract the wave number \(k = 172 \text{ m}^{-1}\) and the angular frequency \(\omega = 4830 \text{ s}^{-1}\). These will be used to calculate the wave speed and further properties.

Step 2: Calculate Wave Speed

The wave speed \(v\) can be calculated using the relation \[v = \frac{\omega}{k}\]Substituting the given values:\[v = \frac{4830 \text{ s}^{-1}}{172 \text{ m}^{-1}} = 28.08 \text{ m/s}\]

Step 3: Determine Time for Pulse to Travel Full Length (a)

Using the wave speed \(v = 28.08 \text{ m/s}\), and string length \(L = 1.50 \text{ m}\):The time \(t\) for a pulse to travel the full length is given by \[t = \frac{L}{v}\]\[t = \frac{1.50 \text{ m}}{28.08 \text{ m/s}} \approx 0.0534 \text{ s}\]

Step 4: Calculate the Tension and Weight W (b)

The wave speed is related to the tension \(T\) and linear mass density \(\mu\) by \[v = \sqrt{\frac{T}{\mu}}\]First, calculate \(\mu\):\[\mu = \frac{\text{weight of string}}{\text{length of string}} = \frac{0.0125 \text{ N}}{9.81 \text{ m/s}^2 \times 1.50 \text{ m}} \approx 0.000850 \text{ kg/m}\]Rearranging for \(T\):\[T = v^2 \cdot \mu = (28.08 \text{ m/s})^2 \times 0.000850 \text{ kg/m}\]\[T \approx 0.670 \text{ N}\]Assuming \(T = W\), \(W \approx 0.670 \text{ N}\).

Step 5: Calculate Number of Wavelengths on String (c)

The wavelength \(\lambda\) is given by the formula:\[\lambda = \frac{2\pi}{k}\]\[\lambda = \frac{2\pi}{172} \approx 0.0365 \text{ m}\]Number of wavelengths \(n\) on the string:\[n = \frac{L}{\lambda}\]\[n = \frac{1.50 \text{ m}}{0.0365 \text{ m}} \approx 41.1\]

Step 6: Determine Equation for Waves Traveling Down (d)

The equation for waves traveling down the string is derived by changing the sign of \(kx\) in the wave equation:\[y(x, t) = (8.50 \text{ mm}) \cos(-172 \text{ m}^{-1} x - 4830 \text{ s}^{-1} t)\]

Key Concepts

Wave speed

Wave speed is essential in understanding how fast waves travel along a string. It is a measure of how far a crest or a trough of the wave moves in a unit of time. The calculation of wave speed in a string involves both the angular frequency and the wave number.
For the given wave equation:

• Angular frequency (\(\omega\)): measures how fast the wave oscillates in time.
• Wave number (\(k\)): measures the number of wave cycles per unit distance.

The formula to find the wave speed (\(v\)) is \[v = \frac{\omega}{k}\].
Substituting the values: \(\omega = 4830\, \text{s}^{-1}\) and \(k = 172\, \text{m}^{-1}\), we find \(v = \frac{4830}{172}\approx 28.08\, \text{m/s}\).
This speed indicates how quickly the wave propagates through the string, crucial for problems involving wave travel time in stringed instruments and mechanics.

Tension in strings

Tension in a string is a primary factor that influences the speed of wave propagation. It arises from the forces acting along the string, which in turn affect the wave characteristics. In this problem, the tension is assumed equal to the weight supported by the string, denoted as \(W\).
The relationship involving tension (\(T\)), wave speed (\(v\)), and linear mass density (\(\mu\)) is captured by the formula: \[v = \sqrt{\frac{T}{\mu}}\].
To find \(\mu\), the linear mass density, we divide the weight of the string by the gravitational force times its length: \[\mu = \frac{0.0125\, \text{N}}{9.81\, \text{m/s}^2 \times 1.50\, \text{m}} \approx 0.000850\, \text{kg/m}\].
With \(v\approx 28.08\, \text{m/s}\), calculate the tension: \[T = v^2 \times \mu = (28.08)^2 \times 0.000850 \approx 0.670\, \text{N}\].
This result informs us about the force needed to maintain given wave characteristics in the string.

Wavelength calculation

Wavelength refers to the distance between consecutive points of the same phase, such as crest to crest, on a wave. It is vital for understanding wave behavior, especially in standing waves on strings, like those in musical instruments.
The formula for calculating the wavelength (\(\lambda\)) from the wave number (\(k\)) is: \[\lambda = \frac{2\pi}{k}\].
In this particular problem, \(k = 172\, \text{m}^{-1}\), thus \[\lambda = \frac{2\pi}{172} \approx 0.0365\, \text{m}\].
This wavelength helps determine how many wave cycles fit along the length of the string, calculated by dividing the total string length by the wavelength. The number of wavelengths (\(n\)) thus is: \[n = \frac{1.50\, \text{m}}{0.0365\, \text{m}} \approx 41.1\].

Wave equation

The wave equation is a mathematical description of the wave's displacement over time and space. It encapsulates how a wave propagates along a string, describing its moving shape. The given equation \([y(x, t) = (8.50\, \text{mm}) \cos(172\, \text{m}^{-1} x - 4830\, \text{s}^{-1} t)]\) provides the wave's profile as a function of position \(x\) and time \(t\).
This type of equation is essential to understanding wave behavior, allowing prediction of the wave's position at any given time. To find the equation for waves moving in the reverse direction, swap the sign of the term with \(x\). The resulting equation: \[y(x, t) = (8.50\, \text{mm}) \cos(-172\, \text{m}^{-1} x - 4830\, \text{s}^{-1} t)\] highlights how wave equations can represent waves traveling in various directions.