Problem 63 - University Physics with Modern Physics

Question

A rubber band of mass \(0.21 \mathrm{~g}\) is stretched between two fingers, putting it under a tension of \(2.8 \mathrm{~N}\). The overall stretched length of the band is \(21.3 \mathrm{~cm} .\) One side of the band is plucked, setting up a vibration in \(8.7 \mathrm{~cm}\) of the band's stretched length. What is the lowest frequency of vibration that can be set up on this part of the rubber band? Assume that the band stretches uniformly.

Step-by-Step Solution

Answer

Answer: The lowest frequency of vibration is approximately 56.11 Hz.

Step 1: Convert units

First, convert the given measurements to SI units. This means we need to convert the mass of the rubber band to kilograms and the lengths to meters: Mass: \(0.21 \mathrm{~g} = 0.00021 \mathrm{~kg}\) Overall stretched length: \(21.3 \mathrm{~cm} = 0.213 \mathrm{~m}\) Vibrating length: \(8.7 \mathrm{~cm} = 0.087 \mathrm{~m}\)

Step 2: Calculate the linear mass density

To obtain the linear mass density \(\mu\), find the mass per unit length of the rubber band by dividing the mass by the overall stretched length: \(\mu = \frac{mass}{length} = \frac{0.00021 \mathrm{~kg}}{0.213 \mathrm{~m}} = 9.859 \times 10^{-4} \mathrm{~kg/m}\)

Step 3: Calculate the fundamental frequency

Now we can use the formula for calculating the fundamental frequency for a vibrating string: \(f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\) Substitute the values we found earlier for the vibrating length, tension, and linear mass density: \(f=\frac{1}{2(0.087 \mathrm{~m})}\sqrt{\frac{2.8 \mathrm{~N}}{9.859 \times 10^{-4} \mathrm{~kg/m}}} = \frac{1}{0.174\mathrm{~m}}\sqrt{\frac{2.8\mathrm{~N}}{9.859 \times 10^{-4} \mathrm{~kg/m}}} = 56.11 \mathrm{~Hz}\) The lowest frequency of vibration that can be set up on this part of the rubber band is approximately \(56.11 \mathrm{~Hz}\).

Key Concepts

Linear Mass Density

When we talk about strings or rubber bands in physics, we're often interested in their 'linear mass density.' This is a measure of mass per unit length and is usually denoted by the symbol \( \mu \). For example, if you have a string that's 2 meters long and has a mass of 6 grams, converting the mass to kilograms (since SI units are standard in physics), we get \( 0.006 \) kg. The linear mass density would then be \( \mu = \frac{0.006 \mathrm{kg}}{2 \mathrm{m}} = 0.003 \mathrm{kg/m} \).

In our exercise, by dividing the mass of the rubber band by its stretched length, we are able to calculate the linear mass density. This property is crucial because it directly affects the vibration of the string. The greater the mass per unit length, the lower the frequency for a given tension.

Vibrating String

Picture a guitarist plucking a string. The string vibrates and makes a sound, right? That vibration happens in what we call 'modes'. The simplest mode of vibration, where the string is moving up and down in a single segment, is known as the fundamental mode, and it produces the fundamental frequency of the string.

The formula to find this fundamental frequency is \( f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \), where \( L \) is the length of the string that's vibrating, \( T \) is the tension in the string, and \( \mu \) is the linear mass density we just talked about. The fundamental frequency is the lowest and usually the loudest one you hear. Higher modes of vibration result in higher frequencies, which are multiples of this fundamental frequency, also known as harmonics.

Tension in Physics

Tension is a force conducted through strings, cables, and other similar objects when they are pulled tight by forces acting from opposite ends. It's like tug-of-war with a rope: the force you feel in the rope is tension. In physics, we usually denote this force with the letter \( T \), and it is measured in Newtons (N) in the International System of Units (SI).

In our exercise, the tension in the rubber band is what's creating the pulling force that allows it to vibrate when plucked. The frequency of the vibration is partly dependent on the tension: the higher the tension, the higher the frequency for a given linear mass density. This is why guitarists tune their strings by adjusting the tension to get the right note, or frequency.