3
$\begingroup$

In my book, an experiment to demonstrate resonance if given with the help of tuning force.

The book suggests the use of two identical tuning forks on two separate sound boxes such that their open end face each other.

The reason given for the use of sound boxes is to produce vibrations of large amplitude because of large surface area of air in the sound box.

My question is since amplitude is basically the energy possessed by the wave, shouldn't it be constant and decrease instead of increasing in the sound box. The tuning fork produces some 'x' amplitude. Now when the air molecules are hit, shouldn't they receive less amplitude as the vibrations from the fork are transferred to the air box's surface and then air.

It would be great if someone could explain why the vibrations in the air box are of larger amplitude than the air surrounding the tuning fork.

(P.s I am a high school student and would prefer explanations in layman's language. Thanks)

$\endgroup$

1 Answer 1

2
$\begingroup$

You do not need a sound box; a tuning fork will sound louder if you place its base on a table.

The metal of which the tuning fork is vibrating and those vibrations are transmitted to the sound box / table and make them vibrate in turn. The surface of the sound box / table are in contact with a greater amount of air and so make a greater volume of air vibrate which results in a louder sound being heard.

For most sound boxes this is not resonance but forced vibration. Indeed the sound boxes on musical instruments are manufactured so that no one note produces a larger amplitude sound wave.
If the length of the sound box, which is closed at one end and open at the other, has a length approximately a quarter of a wavelength of the sound in air then it is a resonance effect and a standing wave is produced within the sound box with the result of producing a larger sound.

When the base of a tuning fork is made to touch a table the sound is louder but dies away quicker than if the tuning fork is held in the hand.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.