Question

Imagine that two waves with frequencies 405.4Hz, and 404.6Hz are combined. Express the sum sin405.4t+sin404.6t as the product of two trigonometric functions, and find the longer time-scale over which the signal varies. Give this time period to two decimal places.

Answer 1

\[\sin405.4t+\sin404.6t= 2\sin405t⋅\cos0.4t \]

how do i find the time period?

Answer 2

hmmmm.... when I see "time period" I think of T = 2 pi divided by omega, where omega is the angular frequency.
This post LOOKS as though its about BEAT FREQUENCIES vibrations of "nearly equal" frequencies.
Taking (yet another) guess at it, 0.4t is the thing which pops out from "mixing" the two higher frequencies. That sort of makes my omega = 0.4 radians per second. That makes my TIME PERIOD for the "popping out" or MIXED frequency 2 pi over 0.4 seconds. 2pi is about 6. So, 2 pi over omega would be about 6 divided by 0.4. that's 60/4 that''s 15 seconds ? WAAAAA ooooo WAAAAA ooooo WAAAAA type of thing were they to be sound waves.
Have I guessed correctly ???? (I'll hide now)
bon chance et bon voyage
http://perendis.webs.com
(ps, in answer to the question's first "instruction", I'll prob try not to imagine it ... - maybe I just have ...) I won't dare try to draw the diagrams ... too many wavy lines.

Answer 3

To find a compound trig function just plug into a sum-to-product trig identity like this one:

\(\sin \theta + \sin \varphi =2\sin \left({\frac {\theta + \varphi }{2}}\right)\cos \left({\frac {\theta - \varphi }{2}}\right)\)

Thing is though, your sinusoid look wrong :-(

If they really are f = 405.4Hz and f =404.6Hz then the equation is

\(\sin 2 \pi (405.4) t + \sin 2 \pi (404.6) t\)

so in form \(\sin \omega t\) ie \(\omega = 2 \pi f = \dfrac{2 \pi }{T}\) so that every cycle or period is re-scaled to \(2\pi \) in a sine curve

To find the period of the combined function you need to see where, after starting at t= 0, the two waves meet again in sync ie are in phase again.



This drawing is for sinusoids with periods \(T_1 = 0.5\) and \(T_2 = 3\). They meet after 6 oscillations of the higher frequency one, not surprisingly.


The purple curve is their combination and you can see the shape over 1 of it's wavelength.

The trick here is to compare periods. So in the example given:


\(\dfrac{T_1}{T_2} = \dfrac{0.5}{3} = \dfrac{1}{6}\) so they re-connect at \( 6T_1 = T_2\)


you can actually turn your numbers into a fraction in the same way.

Answer 4

can anyone help me please.