help SHM :( (1 Viewer)

QuLiT

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II) x double dot = -4(x-3)

there for n^2 = 4

n = 2

therre fore period = 2pi/2
= pi

iii) i would use auxilary angle method on x then differentiate

iv) let xdot = 2 and work it out

edit: should put xdot = plus or minus 2
 

hollyy.

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QuLiT said:
II) x double dot = -4(x-3)

there for n^2 = 4

n = 2

therre fore period = 2pi/2
= pi

iii) i would use auxilary angle method on x then differentiate

iv) let xdot = 2 and work it out

edit: should put xdot = plus or minus 2
huh!?
 

tommykins

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ii) period = 2pi/n = 2pi/2 = pi sec
ii) x' = 2sqrt3cos2t + 2sin2t = Acos(2t-@)
A = sqrt (2sqrt3² + 2²) = sqrt 16 = 4

Acos(2t-@) = 4[cos2t.cos@ + sin2tsin@]

Using 4sin@ = 2
sin@ = 1/2
@ = pi/6

Thus x' = 4cos(2t-pi/6)

iii) for 0<t<pi
4cos(2t-pi/6) = +-2
cos(2t-pi/6) = +-1/2

I'm sure you can solve it from there.
 

QuLiT

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mighta got the name wrong, its where you change asin@ - bcos@ into Ccos( @-alpha) where alpha is the initial phase
 

tommykins

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It is called the auxillary angle method.
 

hollyy.

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QuLiT said:
mighta got the name wrong, its where you change asin@ - bcos@ into Ccos( @-alpha) where alpha is the initial phase
is that the thing that u can use in trig equations instead of the t method?
 

QuLiT

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which text book do you have? il tell you the page no
 

QuLiT

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omg the only text book i don't have lol -.- neway its near the part with general solutions of trignometric equations
 

QuLiT

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here:
Transformations (equations involving the sum of sin and cos / axillary angle method)

  • asinx + bcosx = rsin(x+α)
  • asinx - bcosx = rsin(x-α)
  • acosx - bsinx = rcos(x+α)
  • acosx + bsinx = rcos(x-α)
Where:
  • r = √(a<sup>2</sup> + b<sup>2</sup>)
  • α is in the first quadrant such that tanα = b/a
taken from the maths formula wiki thing
 

dwarven

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bottom half of P281 in the first book. [the green one]
=]
 

hollyy.

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QuLiT said:
here:
Transformations (equations involving the sum of sin and cos / axillary angle method)
  • asinx + bcosx = rsin(x+α)
  • asinx - bcosx = rsin(x-α)
  • acosx - bsinx = rcos(x+α)
  • acosx + bsinx = rcos(x-α)
Where:
  • r = √(a<sup>2</sup> + b<sup>2</sup>)
  • α is in the first quadrant such that tanα = b/a
taken from the maths formula wiki thing
your my hero :)
my teacher called this the R method - because theres and R in it lol.
never heard of axillary
thankyou!
 

tommykins

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回复: Re: help SHM :(

QuLiT said:
here:
Transformations (equations involving the sum of sin and cos / axillary angle method)

  • asinx + bcosx = rsin(x+α)
  • asinx - bcosx = rsin(x-α)
  • acosx - bsinx = rcos(x+α)
  • acosx + bsinx = rcos(x-α)
Where:
  • r = √(a<sup>2</sup> + b<sup>2</sup>)
  • α is in the first quadrant such that tanα = b/a
taken from the maths formula wiki thing
wow, @ is from tan@ = b/a? that's neat.
 

QuLiT

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theres also another way you can do @ lol

you can do cos-1 (a/R)
 

danz90

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QuLiT said:
here:
Transformations (equations involving the sum of sin and cos / axillary angle method)
  • asinx + bcosx = rsin(x+α)
  • asinx - bcosx = rsin(x-α)
  • acosx - bsinx = rcos(x+α)
  • acosx + bsinx = rcos(x-α)
Where:
  • r = √(a<SUP>2</SUP> + b<SUP>2</SUP>)
  • α is in the first quadrant such that tanα = b/a
taken from the maths formula wiki thing
do we actually have to remember like each transformation.. or will they always specify in an exam to which form they want you to change it to?
 

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