An ideal vibrating string will vibrate with its fundamental frequency and all harmonics of that frequency. The position of nodes and antinodes is just the opposite of those for an open air column.
The fundamental frequency can be calculated from
where
T = string tension
m = string mass
L = string length
and the harmonics are integer multiples.
Higher Harmonics:
Higher harmonics within the harmonic series come from successively adding nodes (fixed points, where the string doesn't move) to the standing wave pattern. Every time there is an additional node, the frequency gets higher. The next frequency after the fundamental is known as the second harmonic. Instead of just the two nodes at the places where the string is held, we have added a third node, right in the middle of the string. The standing wave pattern at an instant in time now has half the string moving downward while the other half moves upward. At later times, this pattern reverses. There is both a crest and a trough at any instant in time. This means that the wavelength of the second harmonic equals the length of the string.
We can again find the frequency of the second harmonic, by using the relationship between wave speed, wavelength and frequency. As with all wave phenomena, the wave speed does not change with the frequency. It depends on the properties of the medium, alone. For the second harmonic
In words, the second harmonic has twice the frequency of the fundamental. Since the wave speed is the same for both standing waves, it also follows that the second harmonic has half the wave length as the fundamental. The higher harmonic standing waves are called overtones. The second harmonic overtone can be easily heard on a guitar by laying your finger lightly on the string at the midpoint between the two frets after the string has been plucked. If you do this, you hear a faint, higher frequency tone.
Successively higher harmonics are formed by adding successively more nodes. The third harmonic has two more nodes than the fundamental, the nodes are arranged symmetrically along the length of the string. One third the length of the string is between each node. The standing wave pattern is shown below. From looking at the picture, you should be able to see that the wavelength of the second harmonic is two-thirds the length of the string.
We find the wavelength of the third harmonic from the standing wave pattern shown above; it is two-thirds of the length of the string. We find the frequency of this mode:
Harmonic Series:
The rules for the pattern are
When we somehow transfer energy to a string through a vibration, the harmonic series serves to filter out those vibrations occurring at the special frequencies, n f1. The vibration of the string is a superposition of standing waves, each with a different amplitude.
1. The harmonics formed in air column in an organ pipe open at both ends are
Solution:
both odd and even
2. The harmonics formed in air column in an organ pipe closed at one end are
Solution:
only odd
3. A stretched string is used in musical instruments because
Solution:
It is rich in harmonics
1. The frequency of a stretched uniform wire of certain length is in resonance with the fundamental frequency of closed tube. If length of wire is decreased by 0.5m, it is in resonance with first overtone of closed pipe. The initial length of wire is
Solution:
\({{n}_{1}}=\frac{V}{4L},\ {{n}_{2}}=\frac{3V}{4L},\ and\ {{n}_{1}}{{l}_{1}}={{n}_{2}}{{l}_{2}}\)
= 0.5 m
2. An open pipe resonates to a frequency \({{f}_{1}}\) and a closed pipe resonates to a frequency \({{f}_{2}}\). If they are joined together to form a longer tube, then it will resonate to a frequency of (neglect end corrections)
Solution:
\({{f}_{1}}=\frac{v}{2{{l}_{1}}};\,{{f}_{2}}=\frac{v}{4{{l}_{2}}};\,f=\frac{v}{4\left( {{l}_{1}}+{{l}_{2}} \right)}\)
= \(\frac{{{f}_{1}}{{f}_{2}}}{2{{f}_{2}}+{{f}_{1}}}\)
1 . A uniform string of length L and mass M is fixed at both ends under tension T. Find the frequency with which the string vibrates?
Solution:
Frequency of a vibrating string is given by:
\(f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}\)
\(f=\frac{1}{2L}\sqrt{\frac{T}{{}^{M}/{}_{L}}}\,\,=\,\,\frac{1}{2}\sqrt{\frac{T}{ML}}\)
2. When a closed pipe is suddenly opened than the second overtone of closed pipe and first overtone of open pipe differ by 100 Hz. The fundamental frequency of closed pipe will be
Solution:
Second overtone of closed pipe = \(\frac{5\,\,v}{4\,\,L}\)
First overtone of open pipe = \(\frac{v}{L}\)
But \(\frac{5v}{4L}\,\,-\,\,\frac{v}{L}\,\,=\,\,100\), \(\frac{v}{4L}\,\,=\,\,100\), v = 100 Hz
\ Fundamental frequency = \(\frac{v}{4L}=100\text{Hz}\)
1. Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. Find the shortest distance from the wall at which the air particles have maximum amplitude of vibration. [v = 330 m/s]
Solution:
Since wall is rigid, so at the wall there will be node. Now the point where amplitude is maximum is an antinode. Hence minimum distance from the wall where displacement is maximum is the distance between node and antinode i.e. \(\frac{\lambda }{4}\). So
v = 330 m/s v = 660 Hz
\(\lambda =\frac{v}{\nu }=\frac{330}{660}=\frac{1}{2}\)m
Hence required distance = \(\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}=0.125m\)
2. A wire of density p is stretched between two clamps a distance L apart, while being subjected to an extension l. Y is the Young’s modulus of the material of the wire. What is the lowest frequency of transverse vibration of the wire?
Solution:
The Young’s modulus of the wire is given by
\(Y=\frac{stress}{strain}=\frac{T/A}{l/A}=\frac{TL}{Al}\)
\ \(T=\frac{YAl}{L}\) …(1)
Further, \(\text{m=}\frac{\text{mass}}{\text{length}}\text{=}\frac{\text{volume }\!\!\times\!\!\text{ density}}{\text{length}}\)
= area × density = Ap …(2)
The frequency in the fundamental mode is given by
\(\text{n=}\frac{\text{1}}{\text{2L}}\sqrt{\left( \frac{T}{m} \right)}=\frac{1}{2L}\sqrt{\left( \frac{YAl}{L\times A\rho } \right)}\)
or \(n=\frac{1}{2L}\sqrt{\left( \frac{Yl}{L\rho } \right)}\)