SHM (2 Viewers)

[asianese]

Hey guys,

I'm trying to finish the 3u course this holidays so I can concentrate on other stuff during the term.

I'm using Cambridge 3U Yr12 textbook and some other miscellaneous online resources to do motion. I was going ok until I ran into a wall...

How do I prove v^2 = n^2(a^2-x^2)? There doesn't seem to be an explicit proof in Cambridge, nor does the syllabus have it (although it says Note that from the expression for x and v, v2 + n2x2 = a2n2)

(Sorry I don't know where the latex button is?? Anyone?)

I've got to:

x double dot = -n^2x
d/dx (0.5*v^2) = -n^2x
0.5*v^2 = -n^2 * x^2 / 2 + 0.5*C, some constant C
v^2 = -n^2*x^2 +C

WHERE TO INTRODUCE A?!

Thanks for reading this , help is appreciated.

 

[bleakarcher]

a=-n^2*x
d/dx[(1/2)v^2]=-n^2*x
(1/2)v^2=-(1/2)(nx)^2+C
i.e. v^2=-n^2*x^2+2C
At amplitude x=a, the particle/object has velocity 0 right? So when x=a, v=0:
=>0^2=-n^2*a^2+2C
:.2C=n^2*a^2
Hence, v^2=n^2*a^2-n^2*x^2=n^2[a^2-x^2]

 

[bleakarcher]

Note: Amplitude describes maximum displacement from the point of oscillation in SHM.

 

[Peeik]

This is how I was taught, i hope i didnt make any mistakes ....but im sure i didnt because I got the right result

 

[asianese]

OH.....I see! So its a 'stationary point' at x=A so v=0...I see..thank both of you!

 

[Peeik]

OH.....I see! So its a 'stationary point' at x=A so v=0...I see..thank both of you!

Yeah pretty much, although i like to think of a pendulum while understanding SHM because I like my physics