Problem 24 - University Physics with Modern Physics

Question

A light wire is tightly stretched with tension F. Transverse traveling waves of amplitude \(A\) and wavelength \(\lambda_1\) carry average power \(P_{av,1} = 0.400\) 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 \(P_{av,2}\) carried by the waves?

Step-by-Step Solution

Answer

The new average power is 0.100 W.

Step 1: Understand the relation between wavelength and power

For waves stretched along a wire, the average power carried by the waves is related to the tension in the wire, the amplitude of the wave, and the wavelength. The formula for average power \(P_{av}\) is given by:\[P_{av} = \frac{1}{2} \mu v \omega^2 A^2 \]where \(\mu\) is the linear density, \(v\) is the wave speed, \(\omega\) is the angular frequency, and \(A\) is the amplitude.

Step 2: Relate wave speed with tension and wavelength

The wave speed \(v\) on the wire is related to tension \(F\) and linear density \(\mu\) by \(v = \sqrt{\frac{F}{\mu}}\). Furthermore, the wave speed is also related to the wavelength and frequency by \(v = f \lambda\). Since frequency \(f\) can be expressed in terms of angular frequency \(\omega\) as \(f = \frac{\omega}{2\pi}\), substitute back into wave speed, we have:\[v = \frac{\omega \lambda}{2\pi} \]

Step 3: Examine effect of doubling wavelength

If we double the wavelength to \(\lambda_2 = 2\lambda_1\), without changing \(F\), \(A\), or \(\mu\), the wave speed \(v\) remains the same. Thus, the frequency changes as \(f_2 = \frac{f_1}{2}\), and since \(\omega = 2\pi f\), \(\omega_2 = \frac{\omega_1}{2}\).

Step 4: Calculate new power with changed wavelength

Now, substitute into the power expression:\[P_{av,2} = \frac{1}{2} \mu v \left(\frac{\omega_1}{2}\right)^2 A^2 \]Simplify to find the relation between \(P_{av,1}\) and \(P_{av,2}\):\[P_{av,2} = \frac{1}{4} \left(\frac{1}{2} \mu v \omega_1^2 A^2\right) = \frac{1}{4} P_{av,1}\]This shows that when the wavelength is doubled, the average power is reduced to one-fourth.

Step 5: Calculate new average power

Given \(P_{av,1} = 0.400\) W, the new average power is:\[P_{av,2} = \frac{1}{4} \times 0.400 \text{ W} = 0.100 \text{ W}\]

Key Concepts

Wave Speed Calculation

Understanding wave speed is essential when discussing waves on a stretched wire. The speed of a wave, denoted as \( v \), depends on two main factors: the tension \( F \) exerted on the wire and its linear density \( \mu \). This relationship is expressed by the formula:

• \( v = \sqrt{\frac{F}{\mu}} \)

This equation shows that increasing the tension in the wire will increase the wave speed if the linear density remains unchanged. Conversely, higher linear density will reduce wave speed if the tension stays constant.
This relationship is crucial because wave speed \( v \) helps in determining other wave properties, such as frequency and wavelength. Thus, understanding how tension and linear density affect wave speed is foundational for studying waves.

Wavelength Effect on Power

The wavelength of a wave, represented as \( \lambda \), significantly impacts the power carried by the waves on a wire. Power, specifically average power \( P_{av} \), can be determined using the formula:

• \( P_{av} = \frac{1}{2} \mu v \omega^2 A^2 \)

Here, \( \omega \) is the angular frequency, \( A \) is the amplitude, and \( \mu \) is the linear density.
When the wavelength is altered, it directly influences the frequency \( f \) and angular frequency \( \omega \) of the wave. If the wavelength doubles, like from \( \lambda_1 \) to \( 2\lambda_1 \), the frequency decreases by half as \( f = \frac{\omega}{2\pi} \). Since power is proportional to the square of the angular frequency \( \omega^2 \), halving the frequency results in a quarter of the power.
This means doubled wavelength results in a significant reduction of power, showcasing the profound effect wavelength has on wave energy.

Tension in a Wire

Tension is a key factor in determining the properties of waves traveling through a wire. The term 'tension' refers to the stretching force applied along the wire. It directly influences the wave speed, critical for analyzing wave phenomena.

• Formula for wave speed: \( v = \sqrt{\frac{F}{\mu}} \)

Here, \( F \) stands for the tension in the wire, and \( \mu \) represents the linear mass density.
By increasing tension, the wire becomes capable of supporting faster-traveling waves, affecting their frequency and energy. In this exercise, tension remained constant, demonstrating unchanged wave speed despite wavelength changes.
The unchanging tension showcases a stability in the medium, which is essential for consistency in experiments and calculations involving wave mechanics.