Problem 22 - University Physics with Modern Physics

Question

A piano wire with mass 3.00 \(\mathrm{g}\) and length 80.0 \(\mathrm{cm}\) is stretched with a tension of 25.0 \(\mathrm{N}\) . A wave with frequency 120.0 \(\mathrm{Hz}\) and amplitude 1.6 \(\mathrm{mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

Step-by-Step Solution

Answer

(a) The average power is 0.716 W. (b) The average power becomes 0.179 W when amplitude is halved.

Step 1: Understand the Problem

We need to find the average power of a wave on a wire given its physical properties and wave characteristics. First, we will use the wave speed formula and then the power formula for waves on a string.

Step 2: Calculate Linear Mass Density

The linear mass density (\( \mu \)) is found by dividing the mass of the string by its length: \( \mu = \frac{m}{L} \).Given: \( m = 3.00 \text{ g} = 0.003 \text{ kg} \) and \( L = 80.0 \text{ cm} = 0.80 \text{ m} \).\( \mu = \frac{0.003}{0.80} = 0.00375 \text{ kg/m} \).

Step 3: Calculate Wave Speed

Use the wave speed formula: \( v = \sqrt{\frac{T}{\mu}} \).Given: \( T = 25.0 \text{ N} \) and \( \mu = 0.00375 \text{ kg/m} \).\( v = \sqrt{\frac{25.0}{0.00375}} = 81.65 \text{ m/s} \).

Step 4: Find Average Power of the Wave

Use the power formula: \( P = \frac{1}{2} \mu \omega^2 A^2 v \), where \( \omega = 2 \pi f \) is the angular frequency.Given: \( f = 120.0 \text{ Hz} \) and \( A = 1.6 \text{ mm} = 0.0016 \text{ m} \).Find \( \omega = 2 \pi \times 120.0 = 753.98 \text{ rad/s} \).Calculate power:\( P = \frac{1}{2} \times 0.00375 \times (753.98)^2 \times (0.0016)^2 \times 81.65 = 0.716 \text{ W} \).

Step 5: Determine Effect of Halved Amplitude

If the wave amplitude is halved, the new amplitude \( A' = \frac{1.6}{2} \text{ mm} = 0.0008 \text{ m} \).Power depends on the square of the amplitude (\( A^2 \)), so if we halve the amplitude, the power will become:\( P' = 0.716 \times \left(\frac{0.0008}{0.0016}\right)^2 = 0.716 \times 0.25 = 0.179 \text{ W} \).

Key Concepts

Piano Wire Wave

When examining waves traveling through a piano wire, we are looking at how energy is transferred along a tightly stretched string. A piano wire, which is a type of string used in musical instruments, is a perfect example for studying mechanical waves. In such a system, a wave can travel through the wire, carrying energy that moves with it.

This energy is influenced by factors such as the tension in the wire, its mass, and its length. Here, the vibration of the wire creates a frequency that impacts the tone of the sound produced by the piano. The problem provides us with various parameters, such as the tension in the wire (25 N), frequency (120 Hz), and wave amplitude (1.6 mm), to calculate the wave's behavior. Through this, we can understand how all these elements contribute to the wave's power and energy transmission along the piano wire.

Linear Mass Density

Linear mass density (\( \mu \)) is an important concept when dealing with waves on a string, such as in a piano wire. It essentially represents the mass of the wire per unit length and is a key factor in determining wave behavior.

Here's how it is calculated:

• First, take the total mass of the wire and convert it into kilograms if needed.
• Next, determine the length of the wire in meters.
• Finally, divide the mass by the length to get the linear mass density (\( \mu = \frac{m}{L} \)).

In our problem, the mass is 3.00 g (\( 0.003 \text{ kg} \)) and the length is 80.0 cm (\( 0.80 \text{ m} \)), resulting in a linear mass density of 0.00375 kg/m. Understanding this property helps us evaluate how the mass distribution affects the wave speed and energy traveling through the wire.

Wave Speed Formula

Wave speed (\( v \)) is a crucial factor in determining how efficiently a wave travels along a string. It can be calculated using the wave speed formula: \( v = \sqrt{\frac{T}{\mu}} \). This formula links the tension (\( T \)) in the string and its linear mass density (\( \mu \)) to the speed of the wave.

To use this formula:

• Identify the tension within the string. In our problem, it's set at 25.0 N.
• Calculate the linear mass density, which we previously found as 0.00375 kg/m.
• Substitute these values into the formula to find the wave speed.

For our given values, the wave speed is calculated to be 81.65 m/s. This tells us how quickly disturbances (waves) propagate along the wire, which is directly affected by both the tension and the wire's mass distribution.

Amplitude Effect on Power

The amplitude of a wave (\( A \)) has a significant impact on the power (\( P \)) that the wave carries along a string. Power in the context of waves is the rate at which energy is transferred by the wave. The formula used here is: \( P = \frac{1}{2} \mu \omega^2 A^2 v \), where \( \omega \) is the angular frequency.

Key points about amplitude and power:

• The relation between amplitude and power is quadratic (\( A^2 \)). This means if the amplitude doubles, the power increases by a factor of four.
• Conversely, if the amplitude is halved, the power decreases to one-fourth of its original value.

In the presented problem, when we halve the amplitude from 1.6 mm to 0.8 mm, the power changes from 0.716 W to 0.179 W, demonstrating this relationship. Understanding this helps in appreciating how changes in wave characteristics impact the energy conveyed by the wave significantly.