Problem 70 - College Physics

Question

A university physics professor buys \(100 \mathrm{~m}\) of string and determines its total mass to be \(0.150 \mathrm{~kg} .\) This string is used to set up a standing wave laboratory demonstration between two posts \(3.0 \mathrm{~m}\) apart. If the desired second harmonic frequency is \(35 \mathrm{~Hz},\) what should be the required string tension?

Step-by-Step Solution

Answer

The required string tension is 66.15 N.

Step 1: Calculate linear density

The linear density \( \mu \) is the mass per unit length of the string. Given that the string is 100 meters long and has a mass of 0.150 kg, calculate \( \mu \) using the formula: \( \mu = \frac{0.150}{100} = 0.0015 \; \mathrm{kg/m} \).

Step 2: Use the frequency formula for harmonics

For the second harmonic, the frequency \( f \) is related to the speed \( v \) of the wave and the string length \( L \) by the formula: \( f = \frac{v}{\lambda} \). For the second harmonic, \( \lambda = \frac{L}{2} \) because the second harmonic has two loops in one string length. So, \( v = f \times \lambda = 35 \times 2 \times 3 = 210 \; \mathrm{m/s} \).

Step 3: Use the wave speed formula

The wave speed \( v \) on a string is also given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string. We already calculated \( v = 210 \; \mathrm{m/s} \) and \( \mu = 0.0015 \; \mathrm{kg/m} \).

Step 4: Solve for string tension

Rearrange the wave speed formula to solve for tension \( T \):\[ T = \mu v^2 \]Substitute the values into the equation: \( T = 0.0015 \times (210)^2 \).

Step 5: Calculate the result

Calculate \( T \) as follows:\( T = 0.0015 \times 44100 = 66.15 \; \mathrm{N} \).

Key Concepts

String Tension

String tension is a force that stretches and tightens the string. It is crucial for creating standing waves. In this exercise, we need to find the right amount of tension for a string that produces a specific vibration frequency.
To calculate the required string tension in this case, we use the formula:\[ T = \mu v^2 \]Where:

• \( T \) is the tension in the string
• \( \mu \) is the linear density of the string
• \( v \) is the wave speed

The tension determines how fast the wave travels through the string and affects its overall sound and vibration characteristics.
Finding the correct tension creates the desired frequency, ensuring the standing wave produces the right pitch or note for the second harmonic.

Harmonics

Harmonics, or overtones, refer to specific frequencies at which standing waves are formed on a string. Each harmonic corresponds to a unique wave pattern, creating distinct frequencies or pitches.
In the case of a vibrating string fixed at both ends, like in this exercise, harmonics have distinct nodes and antinodes shaped by the length of the string.For the second harmonic:

• Two loops are formed, with nodes at both ends and one node in between.
• The wavelength \( \lambda \) for the second harmonic is \( \frac{L}{2} \), where \( L \) is the length of the string.
• The frequency rises, doubling that of the fundamental frequency.

Understanding harmonics is important for tuning musical instruments and engineering devices like sensors and speakers.

Wave Speed

Wave speed describes how quickly a wave travels through a medium, such as a string. It is pivotal for understanding how waves behave in various environments. For waves on a string:
Wave speed \( v \) is calculated using the formula:\[ v = f \times \lambda \]where:

• \( f \) is the frequency of the standing wave
• \( \lambda \) is the wavelength of the wave

Wave speed is also determined by the tension and linear density of the string:

• More tension increases wave speed.
• Heavier or denser strings decrease wave speed.

In this exercise, achieving the desired harmonic frequency means the wave speed must be accurately adjusted using both tension and string density.

Linear Density

Linear density, \( \mu \), is a measure of the mass of the string per unit length. It influences how waves propagate along the string, playing a significant role in wave calculations.
For this exercise:The linear density is calculated with:\[ \mu = \frac{m}{L} \]where:

• \( m \) is the total mass of the string
• \( L \) is the total length of the string

A lower linear density results in faster wave speeds, while a higher linear density slows them down.Given a particular tension, knowing the linear density helps in predicting and controlling the behavior of waves, such as setting up the desired frequencies or modes of vibration efficiently.