Problem 71 - Fundamentals of Physics

Question

A transverse sinusoidal wave is generated at one end of a long, horizontal string by a bar that moves up and down through a distance of \(1.00 \mathrm{~cm}\). The motion is continuous and is repeated regularly 120 times per second. The string has linear density 120 \(\mathrm{g} / \mathrm{m}\) and is kept under a tension of \(90.0 \mathrm{~N}\). Find the maximum value of (a) the transverse speed \(u\) and (b) the transverse component of the tension \(\tau\). (c) Show that the two maximum values calculated above occur at the same phase values for the wave. What is the transverse displacement \(y\) of the string at these phases? (d) What is the maximum rate of energy transfer along the string? (e) What is the transverse displacement \(y\) when this maximum transfer occurs? (f) What is the minimum rate of energy transfer along the string? (g) What is the transverse displacement \(y\) when this minimum transfer occurs?

Step-by-Step Solution

Answer

The maximum transverse speed is 7.54 m/s, and maximum tension is 90 N. Both occur at displacement 0.01 m. Maximum energy transfer is 53.496 W at displacement 0 m; minimum transfer is 0 W at displacement ±0.01 m.

Step 1: Calculate Angular Frequency and Wavelength

Given the frequency of the wave is 120 Hz, we can calculate the angular frequency using the formula \( \omega = 2\pi f \). Therefore, \( \omega = 2\pi \times 120 \). \[ \omega = 240\pi \, \text{rad/s} \] The amplitude \( A \) is given as 0.01 m (convert 1 cm to meters).

Step 2: Determine the Wave Speed

To find the wave speed \( v \), we need the linear density \( \mu \) in \( \text{kg/m} \), which is \( 0.12 \text{ kg/m} \). Using the formula for wave speed \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, we get: \[ v = \sqrt{\frac{90}{0.12}} \approx 27.39 \text{ m/s} \] Using the wave speed formula \( v = \frac{\lambda}{T} = f\lambda \), solve for \( \lambda \): \[ \lambda = \frac{v}{f} = \frac{27.39}{120} \approx 0.228 \text{ m} \]

Step 3: Calculate Maximum Transverse Speed (a)

The maximum transverse speed \( u_{max} \) is given by \( u_{max} = \omega A \). Substitute \( \omega = 240\pi \) and \( A = 0.01 \text{ m} \): \[ u_{max} = 240\pi \times 0.01 \approx 7.54 \text{ m/s} \]

Step 4: Calculate Maximum Transverse Component of Tension (b)

The maximum transverse component of the tension \( \tau \) is calculated by using \[ \tau = T \sin(kx - \omega t) \] where \( k = \frac{2\pi}{\lambda} \). The maximum value occurs when \( \sin(kx - \omega t) = 1 \): Thus, \( \tau = T = 90 \text{ N} \).

Step 5: Show Phase Conditions for Maximum Values (c)

For \( \tau \) and \( u_{max} \), the maximum occurs when the sinusoidal term \( \sin(kx - \omega t) = 1 \), which is at an angle of 90 degrees or \( \pi/2 \) radians. At this phase, the transverse displacement \( y = A \). Therefore, \[ y = 0.01 \, \text{m} \].

Step 6: Calculate Maximum Rate of Energy Transfer (d)

The power, or rate of energy transfer, is given by \[ P = \frac{1}{2} \mu \omega^2 A^2 v \] Substituting the known values: \[ \begin{align*} P & = \frac{1}{2} \times 0.12 \times (240\pi)^2 \times (0.01)^2 \times 27.39 \ & \approx 53.496 \, \text{W} \end{align*} \]

Step 7: Conditions for Maximum Energy Transfer (e)

Maximum energy transfer corresponds to the maximum power which occurs when the wave freely propagates, leading to this phase condition at \( y = 0 \) since the power is kinetic energy dominated at zero displacement.

Step 8: Calculate Minimum Rate of Energy Transfer (f)

Minimum energy transfer occurs when the transverse velocity is zero, corresponding to the nodes of the wave. Therefore, minimum power is \( P_{min}=0 \text{ W} \).

Step 9: Conditions for Minimum Energy Transfer (g)

Minimum energy transfer corresponds to conditions at antinodes where the actual transverse velocity is zero, meaning \( y = \pm A \), therefore the displacement at these points is \[ y \approx \pm 0.01 \, \text{m} \]

Key Concepts

Transverse Waves

Transverse waves are a type of wave where the particle displacement is perpendicular to the direction of wave propagation. Imagine shaking one end of a rope up and down; this creates a wave that travels horizontally while the rope itself moves vertically.

• Each particle in the medium moves in a direction perpendicular to the wave's forward motion.
• Common examples include light waves and waves on strings, surfaces, or water.

In the given problem, a transverse wave travels through a string, and its features are characterized by its amplitude, speed, and tension. Understanding how these elements interact can help in comprehending how energy and motion are transferred across the wave.

Angular Frequency

Angular frequency is a measure of how quickly the wave oscillates in terms of angles and is represented by the symbol \(\omega\). It is calculated using the formula \(\omega = 2\pi f\), where \(f\) is the frequency in Hertz.
In the context of our problem, the wave on the string oscillates 120 times per second, making \(\omega = 2\pi \times 120\). This results in \(\omega = 240\pi\) radians per second.

• Higher angular frequency means a quicker oscillation of the wave.
• It helps determine other wave characteristics like maximum speed of particles in the wave.

Knowing the angular frequency allows us to explore how fast energy is transferred across the wave, which is crucial in calculating other wave parameters.

Wave Speed

Wave speed refers to how fast a wave passes through a medium, and it depends on the medium's properties, such as tension and density. The formula used is \(v = \sqrt{\frac{T}{\mu}}\), where \(T\) is the tension and \(\mu\) is the linear density of the string.
Given the tension of 90 N and density of 0.12 kg/m (converted from grams/meter to kg/m), we find the wave speed to be approximately 27.39 m/s.

• Wave speed affects the wavelength and frequency relationship, \(v = f\lambda\).
• Understanding wave speed is crucial for comprehending wave behavior in different media.

It determines how quickly energy and information travel along the wave, vital for solving problems related to dynamic systems like waves on a string.

Energy Transfer in Waves

Energy transfer in waves is an essential concept, illustrating how energy is transported from one location to another through wave motion. This happens without the actual transport of matter. For transverse waves, energy is mainly transferred through the wave's kinetic and potential energy.

• The power of a wave gives an idea of its energy transfer rate, calculated using \(P = \frac{1}{2} \mu \omega^2 A^2 v\).
• In our problem, the maximum power is roughly 53.496 watts when the wave is at maximum amplitude.

Understanding energy transfer helps us determine points of maximum and minimum energy flow in the wave. Energy transfer tells us how waves perform work, carry energy over distances, and how this energy can be harnessed or dissipated as the wave propagates.