Question

A rubber band of mass \(0.4245 \mathrm{~g}\) is stretched between two fingers. The overall stretched length of the band is \(20.91 \mathrm{~cm} .\) One side of the band is plucked, setting up a vibration in \(8.117 \mathrm{~cm}\) of the band's stretched length. The lowest-frequency vibration that can be set up on this part of the rubber band is \(184.2 \mathrm{~Hz}\). What is the tension in each side of the rubber band? Assume that the band stretches uniformly.

Short answer

Answer: The tension in each side of the rubber band is approximately 1.881 N.

Step-by-step solution

Find the linear mass density of the rubber band

First, we need to determine the linear mass density, which is the mass per unit length of the rubber band. We can find this by dividing the mass of the rubber band by its overall stretched length: \(\mu = \dfrac{m}{L_{total}}\) where \(m\) is the mass of the rubber band, and \(L_{total}\) is the overall stretched length. Plugging in given values: \(\mu = \dfrac{0.4245 \mathrm{~g}}{20.91 \mathrm{~cm}}\) Don't forget to convert grams to kilograms and centimeters to meters before calculating the result: \(\mu = \dfrac{0.4245 \times 10^{-3} \mathrm{~kg}}{20.91 \times 10^{-2} \mathrm{~m}} = 2.030 \times 10^{-4} \mathrm{~kg/m}\)

Use the fundamental frequency equation to find the tension

Next, we can use the equation for fundamental frequency to determine the tension in each side of the rubber band: \(f_1 = \dfrac{1}{2L} \sqrt{\dfrac{F}{\mu}}\) In this case, the lowest-frequency vibration is \(184.2 \mathrm{~Hz}\) and the length of the plucked section is \(8.117 \mathrm{~cm}\). Since we want to find the tension in each side of the rubber band, we should divide the length by 2: \(L = \dfrac{8.117 \mathrm{~cm}}{2} = 4.0585 \mathrm{~cm}\) Don't forget to convert the length to meters: \(L = 4.0585 \times 10^{-2} \mathrm{~m}\) Now we can rearrange the equation for tension, \(F\): \(F = 4\mu L^2 f_1^2\) Plug in the known values: \(F = 4(2.030 \times 10^{-4} \mathrm{~kg/m})(4.0585 \times 10^{-2} \mathrm{~m})^2(184.2 \mathrm{~Hz})^2 = 1.881 \mathrm{~N}\) So the tension in each side of the rubber band is approximately \(1.881 \mathrm{~N}\).

Key concepts

Linear Mass Density

Understanding the concept of linear mass density is essential when dealing with the physics of vibrating strings or bands, like our rubber band example. This value represents how much mass exists per unit length of an object, which is crucial for determining how it will respond to forces and vibrations.

In the context of our exercise, to find the linear mass density, we divide the total mass of the rubber band by its entire stretched length. It's important to take care when converting units, ensuring that mass is in kilograms and length in meters for consistency with the SI unit system. Then our formula becomes:
\[\mu = \frac{m}{L_{total}}\]
With \(\mu\) being the linear mass density, \(m\) the mass, and \(L_{total}\) the total length. Correct unit conversion and careful calculations are crucial for accuracy and are the first step in our process to ultimately determine the tension in the rubber band's vibration.

Vibrations in Physics

The study of vibrations in physics typically involves objects that oscillate about an equilibrium position. Vibrations can be simple harmonic movements, as is the case with a plucked rubber band where the vibrations are periodic and sinusoidal.

The lowest-frequency vibration or the fundamental frequency, which we deal with in this example, is the lowest pitch that can be produced by a vibrating object, and it's based on its physical properties, including its mass density and the tension applied to it. The formula involved in our calculation relates the fundamental frequency to the length of the rubber band, the tension, and its linear mass density:
\[f_1 = \frac{1}{2L} \sqrt{\frac{F}{\mu}}\]
Here, \(f_1\) is the fundamental frequency, \(L\) is the length of the segment of the rubber band that is producing the vibration, \(F\) is the tension, and \(\mu\) is the linear mass density. The frequency can indicate how frequently the rubber band vibrates per second, playing a crucial role in the musical pitch of the instrument.

Tension in Strings

The concept of tension in strings, or in our case, a rubber band, refers to the force that is exerted along its length when it's stretched. This force is directly related to how tight or loose the string or band is, and it affects the way vibrations travel through the material.

The tension is vital for calculating the frequency of the vibrations - as we've seen in our problem, where we derived the tension from the fundamental frequency equation. The tension ensures the propagation of the wave (the vibration) along the string, affecting the speed and wavelength, and therefore the frequency of the sound produced. The equation:
\[F = 4\mu L^2 f_1^2\]
allows us to solve for the tension (\(F\)) once we know the linear mass density (\(\mu\)), the length of the vibrating section (\(L\)), and the fundamental frequency (\(f_1\)). Higher tension results in faster vibrations and a higher pitched sound, which aligns with the common experience of tight strings producing higher notes on musical instruments.