Simple Harmonic Motion Question (1 Viewer)

[lyounamu]

The displacement x at time t of a point moving in a straight line is given by x = a sin(nt + @). Find the form which this expression takes if initially
(a) v = 0 and x = -5
(b) x = 0 and the velocity is negative.

Thanks for the help, have not revised this for such a long time.

 

[minijumbuk]

:S
What does "The form which this expression takes" mean? xD

 

[lyounamu]

:S
What does "The form which this expression takes" mean? xD

That basically means, find n, a & the @.

 

[minijumbuk]

Argh, I can't do it either =\
Seems like we're both getting rusty at SHM xD

Let's wait for 3unitz or something =P

 

[lyounamu]

Argh, I can't do it either =\
Seems like we're both getting rusty at SHM xD

Let's wait for 3unitz or something =P

What about 4units? LOL

 

[minijumbuk]

Pfft, 4unitz is noob =P

 

[lyounamu]

I've got few more questions to ask by the way. I am sorry for this:

Assume that over several days of constant weather the cycle of temperatures each day is S.H. between 13 degrees Celsius at 4am and 23 degrees Celsius at 4pm. At what times of the day would the temperature be (a) 18 degrees (b) 15 degrees (c) 21 degrees?

 

[shaon0]

lol i don't know this but i'll have a crack even though i am wrong...
a)8:10 am b)5:20am c)9:20am

 

[ronnknee]

Part a

x = a sin (nt + @)
v = an cos (nt + @)
v² = a² . n² (1 - sin² [nt + @])
v² = a² . n² (1 - [x² / a²])
v² = n² (a² - x²)

When v = 0, x = -5
0 = n² (a² - 25)
Since n² cannot be 0, a² = 25
Therefore a = 5

Therefore x = 5 sin (nt + @) and v = 5n cos (nt + @)

When t = 0, v = 0, x = -5
-5 = 5 sin (@)
Therefore @ = - pi /2

Therefore x = 5 sin (nt - pi/2)



Part b

x = a sin (nt + @)

When t = 0, x = 0
0 = 0 sin (@)

Therefore @ = 0

x = a sin (nt)
v = an cos (nt)

When t = 0
v = na but v is negative
v = -na

Therefore
x = - a sin (nt)


Edit: I just tidied up my working out

 

[lyounamu]

x = a sin (nt + @)
v = an cos (nt + @)
v^2 = a^2 . n^2 (1 - sin^2 [nt + @])
v^2 = a^2 . n^2 (1 - [x^2 / a^2])
v^2 = n^2 (a^2 - x^2)

When v = 0, x = -5
0 = n^2 (a^2 - 25)
Since n^2 cannot be 0, a^2 = 25
Therefore a = 5

Therefore x = 5 sin (nt + @) and v = 5n cos (nt + @)

When t = 0, v = 0, x = -5
-5 = 5 sin (@)
Therefore @ = - pi /2

Therefore x = 5 sin (nt - pi/2)





x = a sin (nt + @)
When t = 0, x = 0
0 = 0 sin (@)
Therefore @ = 0
x = a sin (nt)
v = an cos (nt)
When t = 0
v = na but v is negative
v = -na
Therefore
x = - a sin (nt)

Thanks! I get it now.

 

[minijumbuk]

=O

I got basically the same thing, but didn't you say you need to find 'n' as well? =\

 

[lyounamu]

=O

I got basically the same thing, but didn't you say you need to find 'n' as well? =\

He didn't find n, did he?

I thought we were supposed to find n as well.

EDIT: Just learnt that we didn't need n. I regret spending tonnes of time trying to find it. I blame the poor wording of the question. MAD!!!

 

[ronnknee]

n cannot be found in this question unless another condition is given (eg. when t = 2, x = 4, v = 8 [I made up the numbers]). It's only then we can sub a as well as v and x back into the v² equation and find n

 

[lyounamu]

n cannot be found in this question unless another condition is given (eg. when t = 2, x = 4, v = 8 [I made up the numbers]). It's only then we can sub a as well as v and x back into the v² equation and find n

Well, I thought I was suppoed to find it as the question asked the initial expression of x = a sin(nt + @). It was a really poorly worded question. I was wrestling with the question to find n for 30 minutes and when I looked at the answer, I discovered that there was no value for n in the answer itself. :uhhuh:

EDIT: CAN ANYONE ANSWER MY LATEST QUESTION? No one seems to have noticed it.

 

[ronnknee]

Haha don't worry I noticed it. It just took me a while to draw the solution up

 

[lyounamu]

Haha don't worry I noticed it. It just took me a while to draw the solution up

That's awesome! Thanks. Now, I better start learning Projectile Motion. One down, one to go.

 

[lyounamu]

Assume that the tides rise and fall in S.H.M. A ship needs 10 m of water to pass down a channel safely. At low tide the channel is 9 m deep and at high tide 12 m deep. Low tide is at 9 a.m and high tide at 4 p.m. At what time can the ship safely proceed?

I am slowly forgetting S.H.M. Any help will be appreciated. Thanks!

 

[u-borat]

just remember your general equations.

you know that a=1.5
also period....2pi/n=14hoursX60
so you get a value for n.

i'm too lazy to do the working. xD

using the general equation; x=acos(nt+@)+b, or sin....
let the middle of motion of 10.5 metres=0. (easier that way, as it eliminates b)

when t=0...x=-1.5
therefore. -1.5=1.5cos(nt+@)
therefore, we'll let our basic equation be:
x=-acos(nt+@)
becuase in this way, we eliminate alpha
so
x=-acos(nt)
then plug in required values and solve for t.

done


i think at least.
someone correct me if im wrong; its been a while too.

 

[lyounamu]

just remember your general equations.

you know that a=1.5
also period....2pi/n=14hoursX60
so you get a value for n.

i'm too lazy to do the working. xD

using the general equation; x=acos(nt+@)+b, or sin....
let the middle of motion of 10.5 metres=0. (easier that way, as it eliminates b)

when t=0...x=-1.5
therefore. -1.5=1.5cos(nt+@)
therefore, we'll let our basic equation be:
x=-acos(nt+@)
becuase in this way, we eliminate alpha
so
x=-acos(nt)
then plug in required values and solve for t.

done


i think at least.
someone correct me if im wrong; its been a while too.

Awesome.

 

[lolokay]

for your initial questions - if you know that SHM is the one dimensional motion of an object going around in a circle, you can quickly draw a diagram to find the angle when t=0

in the temperature questions, don't forget that a particular temperature will occur twice in the day