I Mechanical waves-relation between frequency and power

1. Sep 16, 2016

frequency__

so i do understand clearly that the higher the energy of a wave , the higher the amplitude ... but what i was not understanding is how we can create a wave that performs 50 oscillations per second with the same amount of energy that we use to form 1 oscillations per second , with both waves of same amplitude?

2. Sep 16, 2016

Staff: Mentor

3. Sep 16, 2016

frequency__

thank you that helped a lot :D , but i still lack some understanding , is the concept that with the same amount of energy you could create 2 waves with the same properties( amplitude , wave length , ect...) but different frequencies really accurate , because from your helpful answer that allowed me to better understand it , it seems to be not really accurate , thanks in advance :)

4. Sep 16, 2016

Staff: Mentor

You are contradicting yourself. Wave length is the reciprocal of frequency, so you can't hold wave length constant while changing frequency.

Go back and look at the link one more time carefully. Mathematical equations convey the message more clearly than natural language. Study the equations and their implications and you can answer your own question.

5. Sep 16, 2016

frequency__

oh sorry it was a typo , i meant velocity and medium , ect... but thank you i will review it few more times for sure :)

6. Sep 16, 2016

nasu

If you want to have the same energy for the same amplitude but different frequencies you will need to change the parameters of the medium.
If you increase frequency of the waves, in order to keep the energy and amplitude constant you need to change the parameters describing the stiffness and density of the medium. Think about a medium less stiff or less dense (or both). You can intuitively understand that a less stiff medium will result in lower potential energy of the wave and a less dense medium will lover the kinetic energy. Of course, for a wave in medium we usually discuss the energy density.

7. Sep 16, 2016

frequency__

yes after reviewing the link , my friend over there provided i was able to develop a much greater understanding of it , thanks for your answer though ! it was able to mark my point as correct :)