Question

A rubber band of mass \(0.3491 \mathrm{~g}\) is stretched between two fingers, putting each side under a tension of \(1.777 \mathrm{~N}\). The overall stretched length of the band is \(20.27 \mathrm{~cm} .\) One side of the band is plucked, setting up a vibration in \(8.725 \mathrm{~cm}\) of the band's stretched length. What is the lowest-frequency vibration that can be set up on this part of the rubber band? Assume that the band stretches uniformly.

Short answer

Answer: The lowest-frequency vibration that can be set up on this part of the rubber band is approximately \(20.2~\mathrm{Hz}\).

Step-by-step solution

Calculate the linear mass density

First, let's find the linear mass density (\(\mu\)) of the rubber band, which is the mass per unit length. To do this, we will divide the mass by the total stretched length of the rubber band. Since the mass is given in grams and the length is given in centimeters, we will convert the mass to kg and the length to meters. Mass = \(0.3491 \times 10^{-3} \mathrm{~kg}\) Length = \(20.27 \times 10^{-2}\mathrm{~m}\) Now, calculate the linear mass density (\(\mu\)) as follows: \(\mu = \dfrac{mass}{length}\) \(\mu = \dfrac{0.3491 \times 10^{-3} ~\mathrm{kg}}{20.27 \times 10^{-2} ~\mathrm{m}}\)

Determine the vibrating length

Next, we need to find the length of the part of the rubber band that is vibrating. We are given that 8.725 cm of the rubber band's stretched length is vibrating. To use this value in our equation, we need to convert it to meters: Vibrating Length = \(8.725~\mathrm{cm} \times 10^{-2} ~\mathrm{m/cm}= 0.08725~\mathrm{m}\)

Calculate the lowest-frequency vibration

Now that we have the linear mass density and the vibrating length, we can plug these values, and the given tension, into the formula for the fundamental frequency of a vibrating string: \(f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}\) \(f = \dfrac{1}{2 \times 0.08725 ~\mathrm{m}} \sqrt{\dfrac{1.777 ~\mathrm{N}}{\dfrac{0.3491 \times 10^{-3}\mathrm{~kg}}{20.27 \times 10^{-2}\mathrm{~m}}}}\) Calculate the value to find the lowest-frequency vibration: \(f \approx 20.2 ~\mathrm{Hz}\) Therefore, the lowest-frequency vibration that can be set up on this part of the rubber band is approximately \(20.2~\mathrm{Hz}\).

Key concepts

Linear Mass Density

Understanding the linear mass density of a material is crucial when dealing with the physics of vibrating strings. In its simplest form, linear mass density, denoted by the symbol \(\mu\), is a measure of mass per unit length. It is calculated by dividing the total mass of the string by its length after conversion into consistent SI units (mass in kilograms and length in meters).

This property plays a key role in determining how waves propagate through the string. A lower linear mass density means that for the same tension, the string will vibrate at a higher frequency, leading to a higher pitch sound. Conversely, a string with a higher linear mass density will produce a lower pitch.

In the context of our rubber band problem, the mass of the rubber band was provided, and we calculated the linear mass density by dividing this mass by the total length of the stretched rubber band, resulting in \(\mu = \dfrac{mass}{length}\).

Vibrating String

A vibrating string such as a plucked rubber band or a guitar string exhibits fascinating physics. The vibration results from the release of stored potential energy into kinetic energy, which then travels along the string in the form of waves. The portion of the string that is free to vibrate determines which harmonics can form, with the fundamental frequency being the lowest possible frequency of vibration.

In our textbook problem, you'll see that we've focused on the section of the rubber band that was actually vibrating - 8.725 cm of the band's total stretched length. This vibrating length, when converted to meters, allows us to calculate the various possible standing waves and particularly the fundamental frequency; the lowest frequency at which the string can vibrate.

Tension in Strings

The tension in a string is the force stretching it, typically measured in newtons (N). It's a critical factor in determining the frequency of vibration for a string. The greater the tension, the faster the vibrations and, consequently, the higher the frequency of the resulting sound. For strings under the same conditions except for tension, the one under higher tension will produce a note with a higher pitch.

In our rubber band scenario, the tension was uniform and provided as part of the problem. With this information, we could progress to calculate the lowest-frequency vibration of the band using the tension in combination with the linear mass density.

Harmonic Waves

Finally, let's talk about harmonic waves, which are crucial to understanding the behavior of a vibrating string. These waves are the standing wave patterns that form as the string vibrates. The fundamental frequency marks the first harmonic and is the lowest and simplest form of vibration that a string can undergo. It corresponds to the string vibrating in a single segment.

Higher harmonics, also called overtones, involve the string vibrating in multiple segments. These segments are always whole number multiples of the fundamental frequency. For a string fixed at both ends, the wavelength of the fundamental frequency is twice the length of the string. Using the formula \(f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}\), we utilized the principles of harmonic waves to determine the lowest-frequency, or fundamental, vibration that could be set up in the rubber band.