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Kaylis [27]
3 years ago
5

Why does the midi system have the widest range of frequencies

Physics
2 answers:
Zanzabum3 years ago
7 0
<span>Question:
Why does the midi system have the widest range of frequencies

Answer:
</span>
Each of these instruments have a very wide range of notes because these are all treble instruments. Now, if you mean altogether, with a trumpet you can hit a very wide range of notes due to it having such a high octave, same with a violin, piano, clarinet, and organ. Kettle drums just have a very manipulable sound.
But remember it is a MIDI System, and you can adjust the frequency according to your desire; So, honestly, there is no straight answer for this, just how you desire the sound.

I could take an A4 note on a trumpet and stretch the sound waves out to sound like an A1 note just by bending it with a MIDI System. So, the possibilities are endless.

Hope this helps :D

P.S been playing the piano for 7 years so I use midi alot.
Mrac [35]3 years ago
5 0
The Midi System has the widest range of frequencies because it doesn't care about the range of frequency of the actual instrument it is playing. it is a event oriented system.
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Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

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