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Ch.7 - Quantum-Mechanical Model of the Atom

Chapter 7, Problem 87

The speed of sound in air is 344 m>s at room temperature. The lowest frequency of a large organ pipe is 30 s - 1 and the highest frequency of a piccolo is 1.5 * 104 s - 1. Find the difference in wavelength between these two sounds.

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Hey everyone in this example, we need to give the difference in wavelengths between two guitar strings. Where our 6th string Corresponds to a frequency of 82. Hz. and our first string corresponds to a frequency of 329. Hz Were also given the speed of sound or velocity as 344 m/s. Now, because this question is asking us for wavelengths, we want to recall that our formula for wavelength is equal to our velocity divided by our frequency. So we're going to go ahead and calculate for our wavelength for each string. So beginning with our first string Were given a frequency of 329.63 Hz. We should recall that one hurts Is equivalent to one Inverse 2nd. And so we can go ahead and actually interpret this in units of inverse seconds. And that's due to the fact that we're gonna want to take our velocity of the speed of sound and divided by our frequency in inverse seconds. So this is going to allow us to find our wavelength for the first string, Where we have 344 m/s divided by our frequency, Which we interpret as 329.63 inverse seconds. And so now we're able to cancel out our units of seconds with inverse seconds, leaving us with meters as our final unit for wavelength, which is what we want. And this gives us a value of 1.0436 m. Next we're going to find the same value, but for our 6th string. So we're going to interpret our frequency given as 82. hertz in inverse seconds. And we're going to use this to find our wavelength where again we're going to recall that we take our velocity or speed of sound 3 44 m per second and divide that by our frequency, which again we interpret as 82.41 inverse seconds. And so we can cancel out seconds with seconds, leaving us with meters four, our unit of wavelength. And this gives us a value equal to 4.17 m. And so this question wants us to give the difference in these in these wavelengths between our two guitar strings. So we're going to take the higher wavelength at 4.17 m for six string and subtract that from 1.436 m, which was our smaller wavelength from the first string. And so this gives us a difference equal to 4.17 m. And this will actually correspond to choice be in our multiple choice. So this will be our final answer to complete this example as the difference between the wavelengths of our guitar strings. So I hope that everything I went through is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video