Wave Problem -- Total amplitude of fundamental and first three harmonics

  • #1

Homework Statement



  1. An acoustic signal is composed of the first three harmonics of a wave of fundamental frequency 463 Hz. If these harmonics are described, in order, by cosine waves with amplitudes of 0.100, 0.300, and 0.760, what is the total amplitude of the signal at time 0.401 seconds? Assume the waves have phase angles θn = 0.

Homework Equations


F(t)= Sum of Ancos(2nf1t-0n)

The Attempt at a Solution


I simply plugged in the above values into the equation and got 0.00600897, -0.29783, and -0.136345. No idea where to go from here. Any assistance is appreciated!
 

Answers and Replies

  • #2
haruspex
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Homework Equations


F(t)= Sum of Ancos(2nf1t-0n)

The Attempt at a Solution


I simply plugged in the above values into the equation and got 0.00600897, -0.29783, and -0.136345. No idea where to go from here. Any assistance is appreciated!
Not sure how you're getting those numbers. please post full working.
 
  • #3
(0.1)*cos(2pi*463*0.401)=0.00600897
(0.300)+cos(4pi*463*0.401)= -0.29783
(0.760)*cos(6pi*463*0.401)= -0.136345
 
  • #4
haruspex
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(0.1)*cos(2pi*463*0.401)=0.00600897
(0.300)+cos(4pi*463*0.401)= -0.29783
(0.760)*cos(6pi*463*0.401)= -0.136345
Hmmm...
I plugged (0.1)*cos(2*pi()*463*0.401) into OpenOffice Calc and it gives -0.052.
The trouble with computing trig functions of such large angles is that a lot of precision is needed.
I also tried (0.1)*cos(mod(2*pi()*463*0.401;2*pi())) and got the same result.
(In Excel you need to change the semicolon to a comma.)
What are you using for the calculation?
For the other two harmonics I get -0.14 and +0.76.
 
  • #5
rude man
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(0.1)*cos(2pi*463*0.401)=0.00600897
(0.300)+cos(4pi*463*0.401)= -0.29783
(0.760)*cos(6pi*463*0.401)= -0.136345
Ther argument of the cosine function is in radians, not degrees.
Then your numbers will agree with haruspex's.
 

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