More applications of calculus... (1 Viewer)

[FDownes]

I'm stuck on this particular simple harmonic motion question. I can do the first half, but the second has me baffled. Here it is;

a) If x = a sin nt + b cos nt, find the acceleration of the particle in terms of t and show that a = -n²x.

= v = an cos nt - bn sin nt
= a = -an² sin nt - bn² cos nt
= a = -n²(a sin nt + b cos nt)
= a = -n²x

b) Find the amplitude and period of the motion.
= period = (2pi)/n

... and that's as far as I've gotten. There's also a third part to the question;

c) Find the maximum velocity.

Can anyone help me out?

 

[cwag]

hmm...probably wrong...but..

i converted x using the auxilary angle conversion.

x= asin(nt) + 6cos(nt)
x= √(a² + b²) sin (nt + arctan(b/a))

now we have it in the form x = a sin (nt + e)
therefore amplitude = √(a² + b²)
and period = 2pi/n

for velocity

v²=n²(a² - x²)
so v²= n² ( √(a² + b²) - 0)
because x=0 when v is max
so v = n (a² + b²)^1/4)

dunno if im right...seems wrong, i dunno

 

[Affinity]

let R = sqrt(a^2 + b^2)
x = R [ (a/R)sin(nt) + (b/R)cos(nt)]
recognise that a,b,R form a right angled triangle, so
a/R = sin(w) and b = cos(w) for some angle w = arcsin(a/R)

so can rewrite

x = R(sin(w)sin(nt) + cos(w)cos(nt))
= R(cos(nt - w))
should be easier now I hope

beh upstairs beat me to it LOL

 

[conics2008]

just change it into transformation method...

eg R= root of a^2+b^2

tan a = b/a etc etc untill you get it in the form of Rcos,sin(x+a)=