Resonance

Every object produces sound waves with natural frequency. When these sound waves interact with another sound wave with the same frequency, it causes vibration to the wave and begin to produce sound with the same frequency. This phenomenon is called resonance.

Resonance happens when a certain sound of natural frequency causes another object with the same natural frequency to vibrate, producing a louder sound. A very common example of this is the breaking  of a glass. If another sound wave reaches or equals the natural frequency of the glass, let's say, a singer with very high pitch voice, the glass vibrates which makes it to break down. A speaker may also be destroyed if a very loud sound is produced from a source connected to it.

Resonance is also a transfer of energy from one object to another. When a vibrating object tranfers energy to another object, it causes vibration to that object with a frequency similar to that vibrating object.

Using a tuning fork and a tube close on one end and open on the other end, resonance can also be observed. As the tuning fork hits the tube, it produces vibration in the tube which travels through the air inside the tube. Since the tube is closed on the other end, the vibrated waves will just bounce back and forth the tube, meeting up the original wave from the fork, producing standing waves. The formation of standing waves in the tube represents the resonance between the fork and the air inside the column. The air inside the tube vibrates due to the energy transferred by the fork to the tube. This vibration is of larger amplitude, producing a louder sound.

Resonance is also responsible for the production of different sounds or different pitch in musical instruments. Is causes the loudness of the sound produced by the instruments.

Resonance in String Instruments

String instruments are designed as each end of the string are attached. This is to produce standing waves with nodes at each end. A string instrument has its lowest natural frequency called as the fundamental frequency. As a person pluck a guitar string, the action produces vibration on the string. This vibration caused the production of standing waves. The vibration on a string instruments also depends on the position of the hand or object which causes the vibration. The length of vibration can be changed depending on the position of the finger where it distubs the string. The fundamental frequency of a vibrating string can be calculated as

$$f_1=\frac {v}{2L}$$.

Using the formula of waves on a stretched string, $$v=\sqrt {\frac {F}{\mu}}$$, the fundamental frequency can be expressed as

$$f_1=\frac {1}{2L}\sqrt{\frac{F}{\mu}}$$.

Example.

A guitar string has a mass of $$3.18 \times 10^{-3}\;kg$$ and length $$0.81\;m$$. If the speed of wave of the string after vibration is $$134.4\;m/s$$, what is the tension on the string? What is its fundamental frequency?

Given:

$$m= 3.18 \times 10^{-3}\;kg\\ l=0.81\;m\\ v=134.4\;m/s$$

Solution 1.

Solve for the inertial factor of the string.

$$\mu=\sqrt{\frac ml}=\sqrt {\frac {3.18 \times 10^{-3}\;kg}{0.81\;m}}=3.93\times 10^{-3}\;kg/m$$.

Solution 2.

Derive the formula for tension from the general equation of the wave on the stretched string.

$$v=\sqrt {\frac {T}{\mu}} \implies v^2=\frac {T}{\mu} \implies T=v^2 \mu$$.

$$T=(134.4\;m/s)^2(3.93\times 10^{-3}\;kg/m)=71\;N$$

The tension on the string is $$71\;N$$.

Solution 3.

Solve for the fundamental frequency using the equation $$f_1=\frac {1}{2L}\sqrt{\frac{F}{\mu}}$$.

$$f_1=\frac {1}{2(0.81\;m)}\sqrt{\frac{71\;N}{3.93\times 10^{-3}\;kg/m}}=82.97\;Hz$$

The fundamental frequency of the guitar string is $$82.79\;Hz.$$

Resonance in Wind Instruments

Wind instruments are often designed as a tube filled with air and with two open ends. One end - the mouthpiece is for the source of the sound, or where the person playing the instrument blows. The vibration produced in the mouthpiece produces waves which travels through the air column and produces vibration in the air inside. Just like the tuning fork and the tube, standing waves are produced inside the air column causing vibration in the air around the column of the instrument producing sound. The vibration of the lips in the mouthpiece is responsible for the production of different sounds of different frequencies. The frequency can also be adjusted by the covering of the fingers on the holes at the body of the wind instrument - this is to control the exit of air from the air column.

For a pipe with one end closed, the fundamental frequency is expressed as

$$f_1=\frac {v}{4L}$$.

For open-ended wind instruments, the fundamental frequency is

$$f_1=\frac {v}{2L}$$.

Example 2.

A tenor saxophone's length is 0.8$$\;m$$. What is the wavelength? What is its fundamental frequency?

Solution 1.

Since a saxophone is an open pipe, the wavelength is twice the length of the sax $$\lambda=2L$$.

$$\lambda=2(0.8\;m)=1.6\;m$$

Solution 2.

The fundamental frequency can be solved as $$f_1=\frac {v}{2L}$$.

$$f_1=\frac {343\;m/s}{2(0.8\;m)}=214.38\;Hz$$