Problem 24 - University Physics with Modern Physics

Question

A light wire is tightly stretched with tension \(F .\) Trans- verse traveling waves of amplitude \(A\) and wavelength \(\lambda_{1}\) carry average power \(P_{\mathrm{av}, 1}=0.400 \mathrm{W}\) . If the wavelength of the waves is doubled, so \(\lambda_{2}=2 \lambda_{1},\) while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{\mathrm{av}, 2}\) carried by the waves?

Step-by-Step Solution

Answer

The average power of the wave with doubled wavelength is 0.100 W.

Step 1: Understand the Relationship of Power and Wavelength

The average power of a wave on a string is given by the equation \( P_{\text{av}} = \frac{1}{2} \mu \omega^2 A^2 v \), where \( \mu \) is the mass per unit length, \( \omega \) is the angular frequency, \( A \) is the amplitude, and \( v \) is the wave speed. For a wave on a string under tension, wave speed \( v \) is given by \( v = \sqrt{\frac{F}{\mu}} \).

Step 2: Relate Wave Speed and Wavelength

Recognize that the wave speed \( v = f \lambda \), where \( f \) is the frequency and \( \lambda \) is the wavelength. Regardless of wavelength changes, \( v \) remains constant because \( F \) and \( \mu \) are unchanged. Thus, the product \( f \lambda \) must also be constant.

Step 3: Determine Frequency Change

With \( \lambda_{2} = 2 \lambda_{1} \), and knowing \( v = f_1 \lambda_1 = f_2 \lambda_2 \), solve for \( f_2 \): \( f_2 = \frac{f_1 \lambda_1}{2 \lambda_1} = \frac{f_1}{2} \). The frequency halves as wavelength doubles.

Step 4: Calculate New Angular Frequency

The angular frequency \( \omega \) is related to frequency by \( \omega = 2\pi f \). As frequency halves, \( \omega_2 = 2\pi (\frac{f_1}{2}) = \frac{\omega_1}{2} \).

Step 5: Determine New Average Power

Substitute \( \omega_2 = \frac{\omega_1}{2} \) into the power equation: \( P_{\text{av}, 2} = \frac{1}{2} \mu (\frac{\omega_1}{2})^2 A^2 v = \frac{1}{4} \left( \frac{1}{2} \mu \omega_1^2 A^2 v \right) = \frac{1}{4} P_{\text{av}, 1} \).

Step 6: Final Calculation

Since \( P_{\text{av}, 1} = 0.400 \) W, calculate \( P_{\text{av}, 2} = \frac{1}{4} \times 0.400 = 0.100 \) W.

Key Concepts

Wave Speed

Wave speed on a string is an important concept in physics, particularly when studying waves under tension. Simply put, wave speed tells us how fast a wave travels along the string. The formula to calculate wave speed (\( v \) ) is: \[ v = \sqrt{\frac{F}{\mu}} \] where:

• \( F \) is the tension in the string, and
• \( \mu \) is the mass per unit length of the string.

IMPORTANT NOTE: Regardless of changes in wavelength, as long as the tension \( F \) and mass per unit length \( \mu \) remain unchanged, the wave speed \( v \) will also remain constant.
This understanding is crucial when analyzing how other wave properties affect each other, like frequency or power.

Wave Frequency

Wave frequency is essentially how often the wave oscillates in a given second. It's a vital property for determining the behavior of waves.The relationship between frequency (\( f \) ) and wave speed (\( v \) ) can be expressed as:\[ v = f \lambda \] where:

• \( f \) is the frequency,
• \( \lambda \) is the wavelength.

When the wavelength doubles, as it happens in the problem, the frequency changes correspondingly to keep the wave speed constant.
Using the relationship \( v = f_1 \lambda_1 = f_2 \lambda_2 \), we find that \( f_2 = \frac{f_1}{2} \), showing that the frequency halves when the wavelength is doubled.
This illustrates how changes in wavelength directly influence frequency when speed is constant. Understanding this relationship helps in comprehending how waves behave in different mediums or under varying conditions.

Angular Frequency

Angular frequency is another fundamental concept linked closely with wave frequency. It offers a way to express how fast an oscillation occurs in terms of radians.Angular frequency (\( \omega \) ) is calculated using:\[ \omega = 2\pi f \]where

• \( f \) is the frequency.

If the wave frequency changes due to an alteration in wavelength—as shown when it halves—angular frequency will also change.
Specifically, when \( f \) becomes \( \frac{f_1}{2} \), then \( \omega_2 = \frac{\omega_1}{2} \), meaning that angular frequency is directly proportional to the wave frequency. Understanding how angular frequency is related to frequency and how it impacts wave power calculations is crucial. It helps to determine the behavior and characteristics needed to solve more complex physics problems and understand wave dynamics deeply.