help??!?? (1 Viewer)

~*pixie_poo*~

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hello everyone, ive got an assesment on monday on applications of calculus to the physical world, and i am working on the revision sheets we were given, and one question is:
what is the amplitude of
s= 4sint + 3cost
(is it from a SHM question and the answer is 5)
but how do u get the answer??? please help me...
 

nick1048

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the amplitude is 5
Now, remember your transformation method back in prelim trig. days:

r sin (x + @) (where @ represents theta)
r^2 = 3^2 + 4^2
r = 5
x = t (coming from your original equation)

.'. 5 sin (t + @)

By definition

x = a sin (nt + @) (S.H.M by definition)
.'. you just equate co-efficients and get 5 for your amplitude :)
 

nick1048

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ok have u studied transformations??? It's a part of 3u trigonometic equations... if you have Fitzpatrick's book it's on page 46. Read that and my working should make sense. Also write it down on paper because the net symbols can make it confusing. That's all the help i can offer :)
 

FinalFantasy

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~*pixie_poo*~ said:
umm i still dont get it.. sorrry guys
u have:
s= 4sint + 3cost
to find the amplitude, a common method is to get the equation into the form..
x=asin (nt)
and to do that u let
4sin t+ 3 cos t=Asin (t+@)
4 sin t+3 cos t=A [ sin tcos@+cos t sin @]
because sin (A+B)=sinAcosB+cosAsinB

now 4sin t+3 cos t=sin t(Acos@)+cos t(Asin @)
equate coefficients....
4=Acos@ and 3=Asin@
now square both sides of both of these things to get
16=A²cos²@ and 9=A²sin²@
adding them both up... u have
A²cos²@+A²sin²@=16+9
A²(cos²@+sin²@)=25
but cos²@+sin²@=1
.: A²=25
and A=5

now back to ur 4sin t+ 3 cos t=Asin (t+@)
the A there is 5
so u can just conclude that the amplitude is 5

dis is same as wat nick did
 

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