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# Question 5: 2015 AP Physics 1 free response

Fundamental frequencies (first harmonics) of strings.

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• Just curious, why would the end that is moving on the oscillator not be an antinode of the string? Where the string is attached would have to be moving up and down to generate the wave, so then wouldn't it be a standing wave with string length as a fourth of the wavelength?
• This wouldn't happen because the string is fixed to the block on the end (and the oscillator could act as a "hand"), so the wave acts as a fixed end reflection. Because of this, the reflection of the wave, when it gets to the edge of the table, will be on the opposite side than what it originally was on. If the wavelength was four times the string length, the standing wave would be open where the oscillator is (which is what I assume you were getting at with the oscillator being an antinode). The issue with this is that the reflection of the wave then goes back to the oscillator and will hit it when the oscillator is in it's equilibrium (middle) position (due to the relationship between period and wavelength), so it will act as another fixed end that will reflect the wave back on the other side of the string. A situation where the end of the object is an antinode can be seen in closed pipes such as trumpets!