What are forced vibrations and resonance ? Show that only odd harmonics are present in an air column vibrating in a pipe closed at one end. A stretched wire emits a fundamental note of frequency 256 Hz. Keeping the stretching force constant and reducing length of the wire by 10cm, the frequency becomes 320 Hz. Calculate the original length of wire.

Explanation

When sound waves travel down the air column in a cylindrical pipe closed at one end, they are reflected at the closed end with a phase reversal Interference between the incident and reflected waves, under appropriate conditions, sels up atationary waves in the air column with a node at the closed end and an atinode at the open end. Let v be the speed of sound in air. In what follows, we shall ignore the end correction.

In the fundamental mode or first harmonic [Fig. (a)], there is a node at the closed end and an antinode at the open end. The distance between a node and a consecutive antinode is `(lamda)/(4)` where `(lamda)` is the wavelength of sound.
Then `(lamda)`= 4L and the corresponding frequency, `n=(v)/(lamda)=(v)/(4L)` ...(1)
In the next higher mode of vibration, the first overtone, two nodes and two antinodes are formed (Fig (b)]. The corresponding wavelength `lamda_1` and frequency `n_1`, are
`lamda_1=(4L)/(3) and n_1=(v)/(lamda_1)=(3v)/(4L)=3n...(2)`
Thus, the frequency in the first overtone is three times the fundamental frequency. Therefore, the first overtone is the third harmonic.
In the second overtone, three nodes and three antinodes are formed [Fig. (c)]. The corresponding wavelength `lamda_2` and frequency `n_(2)` are
`lamda_2=(4L)/(5) and n_2=(v)/(lamda_2)=(5v)/(4L)=5_n ....(3)
which is the fifth harmonic.
In general, the frequency of the pth overtone (p 1, 2. 3. ) is `n_p=(2p+1)n` ...(4)
i.e., the pth overtone is the (2p +1) th harmonic.
Thus only odd harmonics are present as overtones.