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A string has a fundamental of 130 Hz. A standing wave on it has 650 Hz.

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A string has a fundamental of $130\ \text{Hz}$ . A standing wave on it has $650\ \text{Hz}$ . Which harmonic is this? How many antinodes and nodes does it have, counting the ends?

Need a reminder? The key ideaShow

A string fixed at both ends, or a pipe open at both ends: $f_n = \dfrac{nv}{2L}$ , every harmonic. A pipe closed at one end: $f = \dfrac{v}{4L}, \dfrac{3v}{4L}, \dfrac{5v}{4L}, \ldots$ , odd harmonics only. Resonance tube: $\lambda = 2(l_2 - l_1)$ and $e = \dfrac{\lambda}{4} - l_1$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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