SHM again (1 Viewer)

[Morgues]

look below...

 

[Morgues]

here is a question from last years hsc....

A particle, whose displacement is x, moves in simple harmonic motion.
Find x as a function of t if acceleration = -4x

and if x = 3 and velocity = 6 root 3 when t = 0



if you go down to 4c here http://hsc.ozlpn.com/courses/maths/3u/2001_Maths3_HSC_Joel_s.pdf

there is a solution but is all of that work that he does required? Can't you find x in terms of t by just working out the period and amplitude and then substituting into axos(nt)

 

[Lazarus]

Hmm I remember that question.

I think I did what you said, calculating the period and the amplitude and substituted it in, but I also went further and differentiated that result twice and derived a = -10x from it.

I don't know whether the last step was necessary - I did it just in case.

I certainly don't think you need to go through what Joel did.

The following excerpt is from the Board's exam report:

Many candidates approached this question by integrating the given differential equation
rather than using x = A cos (nt + α). The complex integration and algebra led to few correct
solutions, although some candidates did manage to gain 4 marks. In many of the approaches
taken, the candidates did not check the initial conditions for x and v. Failure to do this
resulted in the loss of one of the available marks.

And from the mapping grid in that report:

1 mark for showing n = 2
1 mark for using t = 0, x = 3
1 mark for differentiating to obtain v and using t = 0, v = -6sqrt(3) to obtain correct second equation
1 mark for evaluation of A
1 mark for evaluation of α

 

[Morgues]

Thanks
I see now, Joel's solution just really confused me and convinced me the question was harder then it really was

 

[yeh]

what do they mean by

1 mark for evaluation of A
1 mark for evaluation of

is that just for finding the values of A and ?

 

[Lazarus]

That's correct... you get a mark for calculating the amplitude and a mark for calculating the phase change.