Simulation · Physics · Class 11
A string fixed at both ends: which notes can it play?
From the lesson Harmonics of strings and pipes in Waves. Change the values and watch what happens.
The idea behind it
NCERT §14.6.1
- A string fixed at both ends must have a node at each end. Only standing waves with a whole number of half-wavelengths fit: L = nλ/2, so λ = 2L/n.
- Its natural frequencies, the normal modes, are ν = nv/2L, n = 1, 2, 3, …, with v = √(T/µ).
- The lowest, n = 1, ν₁ = v/2L, is the fundamental mode or first harmonic. n = 2 is the second harmonic, n = 3 the third, and so on; all whole multiples of ν₁ are present.
- A string usually vibrates in a mix of several modes at once. On a sitar or violin, the point where the string is plucked or bowed decides which modes come out stronger.
- Closed pipe (a glass tube partly filled with water): the closed end is a displacement node, where the pressure change is largest; the open end is a displacement antinode, where the pressure change is least.
- So L = (n + ½)λ/2 and ν = (n + ½)v/2L for n = 0, 1, 2, … The fundamental is v/4L, and the others are 3v/4L, 5v/4L, …: only odd harmonics.
- Open pipe: both ends are antinodes, so L = nλ/2 and ν = nv/2L, n = 1, 2, 3, … Every harmonic is present and the fundamental v/2L is twice that of a closed pipe of the same length.
- If a string or air column is driven at a frequency close to one of its natural frequencies, it resonates. A tabla membrane, clamped all round its rim, has normal modes fixed by the rule that no point on the rim moves.
- Worked example (NCERT): a pipe 30.0 cm long with v = 330 m s⁻¹. Open at both ends, ν₁ = 330/0.6 = 550 Hz and the modes are 550n Hz, so a 1.1 kHz source resonates the second harmonic. Closed at one end, the fundamental is 330/1.2 = 275 Hz and only odd multiples occur; 1.1 kHz is four times 275 Hz, an even multiple, so there is no resonance.
More simulations in Waves
1 more