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BEATS
Orchestrating a Solution
FIRST YOU MUST UNDERSTAND INTERFERENCE
EXPLAINING INTERFERENCEThe interaction between two or more
wave functions
For Example: waves with differing phase constants
+
INTERFERENCEWhen mixing the two wave functions they interfere constructively and destructively in different times and
at different spaces
RESULTANT WAVEThe amplitude heard by you is the resultant wave of the two or more
waves functions interacting
Red: Wave Function 1 Blue: Wave Function 2 Purple: Resultant Wave
Here you can see wave functions
interacting
RESULTANT WAVE II
As you can see the resultant is the sum of the two amplitudes
BEATSBeats:
!Are the resultant variation in amplitude of two or more waves that are interacting
constructively or destructively. !!
However, beats are produced by waves out of frequency
RELATED TO FREQUENCY
The rate at which the amplitude varies is proportional to the frequency difference
*However, if frequency difference is too large than you hear two distinct tones rather than one tone which varies in intensity
EFFECT OF FREQUENCYFunctions out of phase, same frequency
Functions with different frequencies
RESULTANT WAVESResultant wave of functions with same frequency, but different phase
constant
Resultant wave of functions with different frequency
SEE THE DIFFERENCE?Resultant wave of differing phase
constant functions
Resultant wave of differing
frequency functions
SINUSOIDAL EFFECT
Amplitude vs time graph for the sum of the waves
EQUATION FOR RESULTANT AMPLITUDE
Variables: !
ω* = (ω1 + ω2)/2 Mean angular frequency !
Δω = (ω1 - ω2)/2 Angular frequency difference !
t = time !
*set X0 to 0
STotal(0,t) = 2smcos(ω*t)cos(Δωt)
EXAMPLE QUESTION
Exactly 5 minutes into a musical play you hear a peculiar tone that sounds really
good. The sound is made from the combined tones of a guitar and violin.
Being a rockstar, you want to know what tone that was to be able to incorporate
it into your next album. Also being a Physics Major you decide to use your combined knowledge of music and physics to find the tone’s amplitude.
What is the tone’s amplitude if you know that:
The guitar tone has a wave function of : 2cos(4x-3t)
and the violin has a wave function of 2cos(6x-9t)
Sm = nm
SOLUTIONYou want to put the equation in the form
STotal(0,t) = 2smcos(ω*t)cos(Δωt)
You know that S(x,t)= 2cos(4x-3t)
and S’(x,t)= 2cos(6x-9t)
So therefore ω = 3 Hz
and ω’ = 9 Hz
SOLUTIONTherefore find
!ω* = (ω1 + ω2)/2 (Mean angular frequency)
& Δω = (ω1 - ω2)/2 (Angular frequency difference)
ω* = (3+9)/2 = 6 Hz
!Δω = (3-9)/2
= -3 Hz
SOLUTION
At t = 5 minutes t = 5minutes * 60seconds/1minute
t = 300s
STotal(0,t) = 2smcos(ω*t)cos(Δωt)
STotal(0,t) = 2(2)cos(6(300))cos(-3(300))
The tone you heard had an amplitude of -0.263 nm
END OF BEATS LO
Graphing done at Desmos Online Graphing Calculator