Problem with last step of SHM derivation

  • Thread starter Thread starter funcosed
  • Start date Start date
  • Tags Tags
    Derivation Shm
Click For Summary
SUMMARY

The discussion focuses on deriving the position function of a mass oscillating on a spring, described by the equation F = -ky. The solution involves using the characteristic equation to find roots, leading to the general solution x = Aexp(iwt) + Bexp(-iwt) or its real form x = Asin(wt) + Bcos(wt). The user seeks clarification on converting the sine and cosine form into the cosine form with a phase shift, x = Ccos(wt + φ), and receives guidance on the necessary trigonometric identities and normalization of coefficients.

PREREQUISITES
  • Understanding of classical mechanics, specifically Hooke's Law.
  • Familiarity with differential equations and characteristic equations.
  • Knowledge of trigonometric identities and transformations.
  • Basic skills in complex numbers and their applications in oscillatory motion.
NEXT STEPS
  • Study the derivation of the characteristic equation for second-order differential equations.
  • Learn about the application of trigonometric identities in converting between sine and cosine forms.
  • Explore the concept of phase shift in oscillatory motion and its physical significance.
  • Investigate the use of complex exponentials in solving differential equations in physics.
USEFUL FOR

Students of physics, particularly those studying mechanics and oscillatory motion, as well as educators looking for clear explanations of spring dynamics and differential equations.

funcosed
Messages
35
Reaction score
0

Homework Statement


A masss m ossilates on a spring, find the position as a function of time.


Homework Equations


F=ma
F=-ky


The Attempt at a Solution



Well known problem, set
m(dx/dt) = -kx
get roots of characteristic equation and sub into general solution,
x = Aexp(iwt) +Bexp(-iwt)
or
x = Asin(wt) + Bcos(wt)
I'm ok down as far as here however I know this can be written as Ccos(wt-) but I can't see how?
 
Physics news on Phys.org
(1) probably just typo, not mx'=-kx but mx''=-kx

(2) if you want to go from x = Asin(wt) + Bcos(wt) to x=Ccos(wt+phi), you would have to rewrite the first expression as

x = sqrt(AA+BB) (Asin(wt)/sqrt(AA+BB) + Bcos(wt)/sqrt(AA+BB))

( sqrt(a) is squared root from a)

as soon as (A/sqrt(AA+BB))^2+(B/sqrt(AA+BB))^2 = 1, you can say that there is some angle phi for which cos(phi)= B/sqrt(AA+BB), sin(phi)=A/sqrt(AA+BB). Then use formula for cos of difference.
 
thanks
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
2K