Problem 38 - College Physics

Question

A \(\mathrm{A} 75.0 \mathrm{cm}\) wire of mass 5.625 \(\mathrm{g}\) is tied at both ends and adjusted to a tension of 35.0 \(\mathrm{N} .\) When it is vibrating in its sec- ond overtone, find (a) the frequency and wavelength at which it is vibrating and (b) the frequency and wavelength of the sound waves it is producing.

Step-by-Step Solution

Answer

(a) Frequency: 136.12 Hz, Wavelength: 0.5 m. (b) Frequency: 136.12 Hz, Wavelength: about 2.52 m in air.

Step 1: Convert Mass to Kilograms

The mass of the wire is given as \(5.625 \text{ g}\). We need to convert this mass to kilograms for the calculations. Divide the mass by 1000 to convert grams to kilograms:\[m = \frac{5.625}{1000} = 0.005625 \text{ kg}\]

Step 2: Calculate Linear Mass Density

The linear mass density \(\mu\) is the mass per unit length of the wire. The length \(L\) of the wire is \(75.0 \text{ cm} = 0.75 \text{ m}\). Thus, the linear mass density is:\[\mu = \frac{m}{L} = \frac{0.005625}{0.75} = 0.0075 \text{ kg/m}\]

Step 3: Use Tension and Linear Density to Find Wave Speed

The wave speed \(v\) in the wire is found using the formula \(v = \sqrt{\frac{T}{\mu}}\) where \(T = 35.0 \text{ N}\) is the tension in the wire:\[v = \sqrt{\frac{35.0}{0.0075}} \approx 68.06 \text{ m/s}\]

Step 4: Identify the Mode of Vibration

The wire is vibrating in its second overtone. The second overtone is equivalent to the third harmonic. For the third harmonic, the relationship between the wavelength \(\lambda\) and the length \(L\) of the wire is:\[\lambda = \frac{2L}{3} = \frac{2 \times 0.75}{3} = 0.5 \text{ m}\]

Step 5: Calculate the Frequency of Vibration

Using the wave speed formula \(v = f\lambda\), we can find the frequency \(f\):\[f = \frac{v}{\lambda} = \frac{68.06}{0.5} \approx 136.12 \text{ Hz}\]

Step 6: Determine Wavelength and Frequency of Sound in Air

In air, the wavelength of the sound wave produced is the same as that of the vibration in the wire, which is \(0.5 \text{ m}\). The frequency is also the same, \(136.12 \text{ Hz}\), assuming the speed of sound in air is approximately \(343 \text{ m/s}\). To check consistency: the wavelength for this frequency in air would be \(\lambda_{air} = \frac{343}{136.12} \approx 2.52 \text{ m}\), which is not constrained by the wire's length.

Key Concepts

Wave Speed Calculation

Calculating the wave speed in a vibrating string is a crucial part of understanding wave mechanics. Wave speed (\( v \)) is determined by the tension (\( T \)) applied to the string and the string's linear mass density (\( \mu \)). The relationship is given by the formula:

• \[ v = \sqrt{\frac{T}{\mu}} \]

In this context, the tension is the force applied along the string, which in the original exercise is 35 N. This force generates waves throughout the material. By determining the linear mass density (explained in the next section), one can use this formula to find out how fast waves travel along the string.
This is a vital step because knowing the wave speed allows further calculations concerning wave frequency and wavelength.

Linear Mass Density

Linear mass density (\( \mu \)) is a measure of how much mass is distributed per unit length along the string. It is essential for calculating wave characteristics. To find the linear mass density, divide the total mass (\( m \)) of the string by its length (\( L \)):

• \[ \mu = \frac{m}{L} \]

In our exercise, the mass of the wire is converted to kilograms before finding the linear mass density — from 5.625 g to 0.005625 kg. The wire length is measured to be 0.75 m, so \( \mu \) is computed as 0.0075 kg/m.
This value is critical as it contributes directly to calculating the wave speed and indirectly affects the wave's frequency.

Second Overtone

When a vibrating string is said to be in its second overtone, it means it is vibrating in the third harmonic mode. Understanding overtone and harmonics is essential in wave physics. In simple terms:

• The fundamental frequency corresponds to the first harmonic.
• The first overtone corresponds to the second harmonic.
• The second overtone corresponds to the third harmonic.

For the third harmonic, the string length (\( L \)) can hold three half-wavelengths. Therefore, the wavelength (\( \lambda \)) for the second overtone is given by:

• \[ \lambda = \frac{2L}{3} \]

This is a unique feature of standing waves in a fixed medium, where nodes and antinodes form a pattern. Each harmonic has a distinct wavelength and frequency, integral to solving vibrational problems accurately.

Frequency Calculation

The frequency (\( f \)) of a wave signifies how often the wave oscillates per second. Calculating the frequency of a vibrating string involves using the wave speed (\( v \)) and wavelength (\( \lambda \)) with the formula:

• \[ f = \frac{v}{\lambda} \]

In our scenario, with a wave speed of approximately 68.06 m/s and a calculated wavelength of 0.5 m, the frequency is:

• \( f \approx 136.12 \, \text{Hz} \)

This frequency translates directly to the sound wave frequency in air since the frequency remains constant when the medium changes, even though the speed of sound in air is different. Understanding how to determine frequency is critical in applications like music production and acoustic engineering.