Stationary waves on a stretched string
A string fixed at both ends must have a node at each end, so L = n(λ/2). This gives λₙ = 2L/n and fₙ = nv/2L = n f₁, with all harmonics n = 1, 2, 3, … present. The fundamental (first harmonic) is a single loop with λ = 2L.
λₙ = 2L/n · fₙ = nv/(2L) = n f₁ · v = √(T/μ)What this actually means
Start every one of these questions by drawing the loops and marking the boundary conditions. Node at each fixed end, then fit a whole number of loops between them. The algebra falls out of the picture.
All harmonics are allowed here, so the series is f₁, 2f₁, 3f₁, 4f₁ …. Do not carry the closed-pipe odd-only rule across to strings.
Watch the naming. The nth harmonic is the (n − 1)th overtone. The second harmonic and the first overtone are the same thing, and questions switch between the two languages deliberately.
v = √(T/μ) is not on the 9749 formula sheet as a required equation, but you should know that tightening the string or using a lighter string raises v and therefore raises every resonant frequency. Changing frequency does not change v.
Number the loops, not the nodes. A string in its third harmonic has three loops and four nodes including the two ends.
Counting loops as wavelengths, giving λ = L for the fundamental instead of λ = 2L.
Prove it — watch it be true
- Select the 3D string and step the mode selector through n = 1 to n = 5.
- Confirm every integer mode exists and that mode n always shows exactly n loops with a node at each end.
- Check that the frequency readout for each mode is n times the n = 1 value.