Question

Bob is talking to Alice using a tin can telephone, which consists of two steel cans connected by a 20.0 - m-long taut steel wire (see the figure). The wire has a linear mass density of \(6.13 \mathrm{~g} / \mathrm{m},\) and the tension on the wire is \(25.0 \mathrm{~N}\). The sound waves leave Bob's mouth, are collected by the can on the left, and then create vibrations in the wire, which travel to Alice's can and are transformed back into sound waves in air. Alice hears both the sound waves that have traveled through the wire (wave 1 ) and those that have traveled through the air (wave 2), bypassing the wire. Do these two kinds of waves reach her at the same time? If not, which wave arrives sooner and by how much? The speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\). Assume that the waves on the string are transverse.

Short answer

Short Answer: The wave through the wire arrives sooner at Alice's end than the wave through the air. The difference in time between the arrival of the two waves is approximately 0.047 seconds.

Step-by-step solution

Calculate the speed of the wave in the wire

To find the wave speed of the wire, we can use the formula: \(v_{wire} = \sqrt{\frac{T}{\mu}}\), where \(v_{wire}\) is the wave speed, \(T\) is the tension in the wire, and \(\mu\) is the linear mass density of the wire. The problem gives us the tension \(T = 25.0 \thinspace N\) and the linear mass density \(\mu = 6.13 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\). Let's plug these values into the formula: \(v_{wire} = \sqrt{\frac{25.0 \mathrm{~N}}{6.13 \times 10^{-3} \mathrm{~kg} / \mathrm{m}}}\) Next, we'll find the time it takes for each wave to travel 20 meters.

Calculate the time for each wave to travel 20 meters

The time it takes for each wave to travel the 20-meter distance can be calculated using the formula: \(t = \frac{d}{v}\), where \(t\) is the time it takes, \(d\) is the distance traveled, and \(v\) is the speed of the wave. For the wave through the wire, we have: \(t_{wire} = \frac{20.0 \mathrm{~m}}{v_{wire}}\) For the wave through the air, we have: \(t_{air} = \frac{20.0 \mathrm{~m}}{343 \mathrm{~m} / \mathrm{s}}\) Now, we compare the times for both waves and calculate the difference.

Compare the times and find the difference

We need to find out which wave arrives sooner at Alice's end and by how much time. To do this, we just compare the calculated times and find the difference: \(\Delta t = |t_{wire} - t_{air}|\) Let's calculate the values and find the difference.