University of Wisconsin Green Bay

Different musical instruments sound different. In other words, you can distinguish a piano from a trumpet, even if they both play the same note. This is because of the relative strengths of the overtones that are present in the sound waves produced by each. What are the frequencies of the first three overtones for a piano playing a Middle C (262 Hz)? If a piano tuner found that the Middle C string actually had a fundamental frequency of 255 Hz, by what factor would he have to adjust the tension in order to tune the string? Should he increase or decrease the tension?

  • In this problem, you are directly told to consider overtones of vibration. Overtones are standing wave patterns that result from the superposition of identical waves in the same place at the same time. Therefore, this is a 1-dimensional standing wave problem. You can bring in additional definitions as you need them, but always begin with the key physics of the situation.

  • reference frames

    Even though you are asked for the overtones, always begin with a picture of the fundamental oscillation. The fundamental oscillation is the simplest wave pattern that meets the boundary conditions. In this case, you know that the piano strings are anchored at each end and so there are nodes at each end of the string. The simplest pattern that accomplishes this is half a wavelength.

    Once you have the fundamental oscillation, you can build the overtones in order just by making them increasingly complicated. Each addition of a node takes you to a higher order overtone.

  • Standing wave pictures give a picture of, and a way to measure, the the wavelength of the oscillation of interest. In this case, you are asked to consider the frequencies associated with the wavelengths. Wavelength and frequency are always related by the wave equation:

    v = f λ

    If at this point you also recognize that in later questions you will need the definition of the velocity of a wave in a string that is great. If you do not see that part of the problem yet, that is fine. You will come to the additional question as you work the problem. Velocity of a wave in a string is given by

  • We now know the frequencies of the first three overtones. Scroll down to consider how to tune the piano string.


    Step 2

    The problem tells us that the piano tuner adusts the pitch of the note by changing the fundamental frequency, and so we only need to look at the first picture.

    Furthermore, we are told that he adjusts the frequency by adjusting the tension. If you did not previously realize that you need to use the definition of velocity of a wave in a string, you would see it now.

    At this point, you can answer the conceptual part of the question. The frequency of the string's vibration is too low, and so in order to increase the frequency you need to also increase the tension. To quantize the amount by which tension needs to be adjusted, compare the two cases as a ratio.

    The tension in the string needs to be increased by a factor of 1.06, or 6 %, in order to tune the string.

  • In this problem, the word "overtones" should flag that you want to examine standing waves in a string, even if you aren't yet comfortable recognizing the relationship between standing waves, resonance, and musical instruments. All you need in order to understand 1-dimensional standing waves is a picture of the waves and the wave equation.

    That said, if you play a musical instrument make sure to relate your answers to your musical knowledge. You may know that the frequency of a note doubles each octave, for example, and so you may recognize the first overtone as being the C one octave above Middle C. You also know that as you tighten a guitar string the pitch of the string gets higher, in agreement with your conceptual answer to the second part of the problem. You can now explain that effect because you recognize that tightening the string increases the speed of the wave in the string. If wavelength is fixed by the instrument, f must increase so that v = f λ remains true.