Simple harmonic motion (1 Viewer)

[allstarr69]

a quick question which im very lost over:
a particle, whose displacement is x, moves in simple harmonic motion. Find x as a function of t if
x dot dot = -4x

and if x= 3 and x dot = -6root 3 when t = 0

 

[CrashOveride]

u can find the amplitude from v^2 = n^2(a^2-x^2)
then becoz u know its SHM u know it can be of the form x=asin(nt+@)
then solve from ther using the conditions

 

[superbird]

question. is x=acos(nt+@) also an acceptable answer?

 

[Li0n]

yes, the t addition in the brackets just turns out to be different since sin(90-@) = cos@, so it will work out to be the same anyway

 

[paper cup]

question. is x=acos(nt+@) also an acceptable answer?

you need to use the initial conditions to see which form to use.
If particle is originally at origin use sin, if not use cos. I think. I hope...

 

[mojako]

well assuming u can quote x=asin(nt+@) or x=acos(nt+@),
just choose one of them, say
x=asin(nt+@)
but since x dot dot = -4x = -n²x,
n=2

x=asin(2t+@)
x dot or v=2acos(2t+@)

then put t=0 to both, you get
3=asin@ --(1)
-6root3=2acos@ --(2)

Square equation 1 then square equation 2.
add them, and use the fact that sin²@+cos²@=1


Now if you're not allowed to quote x=asin(nt+@),

"x dot dot" = acceleration
= d/dx (1/2 v²)

continue from there by integrating with respect to x and you get v² in terms of x. Then take square root, positive or negative doesn't matter (technically you should do both... but they'll give you the same final result)
use v=dx/dt and invert that (so it becomes dt/dx) and integrate again with respect to x.

 

[Steven12]

follow the above method

the answer is

x=3root3Cos(2t+pie/6)


pieeeee, ummmmmmmmmmmmmmm.............

 

[senso]

Square equation 1 then square equation 2.
add them, and use the fact that sin²@+cos²@=1

I saw this in a proof for a shm question i did today but didn't understand it. What is going on there. When does sin^2@ + cos^2@ come into it?

 

[mojako]

I saw this in a proof for a shm question i did today but didn't understand it. What is going on there. When does sin^2@ + cos^2@ come into it?

(1): 3=asin@
(2): -6root3=2acos@

(1)^2: 9=a^2 sin^2@
(2)^2: 108=4a^2 cos^2@, which becomes 27=a^2 cos^2@


(1)^2 + (2)^2:
9 + 27 = a^2 sin^2@ + a^2 cos^2@
9 + 27 = a^2 (sin^2@ + cos^2@)
9 + 27 = a^2 (1)
a=6 since we define a to be positive
(well you can use the negative if you want, and you'll just get different @)

But you can find a and @ without this squaring and adding. I am just used to it.
Use triangles.
Draw a triangle using equation (1), you get opposite as 3 and hypotenuse as a.
Then add info from equation (2), i.e. you now know adjacent is 3root3.
The angle will be in quadrant 2 though since the sin is +ve and the cos is -ve.
So you can find the angle using tan in the triangle.
Then you can find a now using equation (1) or (2).

 

[ToO LaZy ^*]

got a question...
a particle is moving in a straight line with SHM. the velocity of the particle is respectively rt20ms^-1 and 4ms^-1 at distances of 1 metre and 2 metres from the centre of motion. find the period and amplitude of the motion

 

[mojako]

by inspection(tm):
n=2/root3
T=2pi/n = pi root3
a=4

 

[akira276]

another acceptable answer to qu is x=6sin(2t+pi/6)

i like this form rather than cos fro some reason....

 

[gordo]

you need to use the initial conditions to see which form to use.
If particle is originally at origin use sin, if not use cos. I think. I hope...

if it starts at origin use sin, if it starts at maximum use cos, if it starts a minimum use -cos

 

[Sirius Black]

by inspection(tm):
n=2/root3
T=2pi/n = pi root3
a=4

i got the same answers as you

btw, are we allowed to use the eqn v²=n²*(a²-x²) without proof in the exam?

 

[mojako]

i got the same answers as you

btw, are we allowed to use the eqn v²=n²*(a²-x²) without proof in the exam?

yes u r allowed
but I didn't used it
it'll make it easier though

if you're interested with my method:
it's based on inspection(tm)

go learn now

 

[Sirius Black]

if you're interested with my method:
it's based on inspection(tm)

go learn now

what is it? i still don't get what you said

 

[ToO LaZy ^*]

i got the same answers as you

btw, are we allowed to use the eqn v²=n²*(a²-x²) without proof in the exam?

how did you solve it using that?

 

[ToO LaZy ^*]

what is it? i still don't get what you said

he's just playing with ya..
he worked it out on paper, but only typed the answer and not the working to make you think he did it in his head..

 

[mojako]

what is it? i still don't get what you said

type inspection in google!!!
learn to use search engine

maths pros only do questions based on inspection
and im trying to join the club

although... im definitely not as pro as the pros in this forum
estimated mark of 80-90/120 in ext2 whilst they have about... >115/120
*cry*


well anyway I'll elaborate on my inspection method:
take
x=asin(nt+@) [or cos if you want]
v=ancos(nt+@)

then substitute the given values and you have 4 equations. Square and add pairs of them with suitable selection based on inspection(tm) principle.
Then solve the two resulting equations simultaneouly by suitable substitution, again using the principle of inspection(tm), taking n and a as the positive values.

 

[mojako]

he's just playing with ya..
he worked it out on paper, but only typed the answer and not the working to make you think he did it in his head..

and Sirius Black knows about inspection.
He's just pretending....