Simple Harmonic Question (1 Viewer)

[nofate]

I'm interested in how to solve this question. The way I did it (finding n as 3, integrating acceleration to find v, then finding the amplitude, then creating an equation in the form x = acos(nt + alpha)) seemed too complex and gave me the wrong answer for alpha (i got a positive when it should be negative).

The question is:

A particle moves in a straight line so that its acceleration at any time is given by d²x /dt² = -9x. Find its period, amplitude and displacement at time t if initially the particle is 2 cm from the origin and has velocity 2root3.




If someone is posting a solution, could they post two solutions; one with displacement as sine, the other with displacement as cosine

 

[Sy123]

Very smart question indeed, posting solution.

Amplitude = 4/sqrt3
Period=2pi/3

Also that pi/3 at the end is PART of the Sine function.

 

[Sy123]

Here it is in cosine form, Im absolutely sure Im right.

 

[RishBonjour]

Here it is in cosine form, Im absolutely sure Im right.

Got the same answer. he is definitely right.


You must spread some Reputation around before giving it to Sy123 again.

 

[Timske]

Where does x=sqrt3/2 come from

 

[Sy123]

OOPS sorry, in my line of working
It is supposed to say, t=0 ---> x=2
:/

Then you sub in x=2, and you get root3/2, which simplifies to pi/3, times by 1/3 put to the other side makes C=-pi/9