Question

A rubber band of mass \(0.1701 \mathrm{~g}\) is stretched between two fingers, putting each side under a tension of \(1.851 \mathrm{~N}\). The overall stretched length of the band is \(21.55 \mathrm{~cm} .\) One side of the band is plucked, setting up a vibration in the band's stretched length. The lowest- frequency vibration that can be set up on this part of the rubber band is \(254.6 \mathrm{~Hz}\). What is the length of the vibrating part of the band? Assume that the band stretches uniformly.

Short answer

Answer: The length of the vibrating part of the rubber band is approximately 9.54 cm.

Step-by-step solution

Find the linear mass density of the rubber band

To find the linear mass density (\(μ\)) of the rubber band, we have to use the given mass and length. The linear mass density is given by the formula: \(μ = \frac{M}{L}\) Where \(M = 0.1701 \mathrm{~g}\) is the mass of the rubber band (convert this to kg), and \(L = 21.55 \mathrm{~cm}\) is the stretched length of the band (convert this to meters). \(μ = \frac{0.1701 \times 10^{-3} \mathrm{~kg}}{0.2155 \mathrm{~m}} \approx 0.000789 \mathrm{~kg/m}\)

Determine the wave speed on the rubber band

Now that we have the linear mass density, we can determine the wave speed using the following formula: \(v = \sqrt{\frac{T}{μ}}\) Where \(T = 1.851\mathrm{~N}\) is the tension and \(μ\) is the linear mass density. \(v = \sqrt{\frac{1.851}{0.000789}} \approx 48.526 \mathrm{~m/s}\)

Calculate the length of the vibrating part of the rubber band

Using the equation relating the frequency, wave speed, and vibrating part's length, we have: \(f = \frac{v}{2L}\) Let's solve for \(L\): \(L = \frac{v}{2f}\) Plugging in the given frequency (\(f = 254.6\mathrm{~Hz}\)) and calculated wave speed (\(v = 48.526\mathrm{~m/s}\)), we get: \(L = \frac{48.526}{2 \times 254.6} \approx 0.0954 \mathrm{~m}\) or \(9.54 \mathrm{~cm}\) Hence, the length of the vibrating part of the rubber band is approximately \(9.54 \mathrm{~cm}\).