Question

Can somebody explain me why the following equations are solutions for this differential equation? (Procedure)

Answer 1

Differential equation:
\[\frac{ d ^{2}x }{ dt ^{2} }=-\omega ^{2}x\]

Solutions:
x=Asen(ωt)
x=Acos(ωt)
x=Acos(ωt+θo)
Where A: amplitude, w: angular frequency, t: time and x: position.

Answer 2

take x=Asin(ωt) and put it in the differential...if it satisfies it that means its a solution for the d.e

Answer 3

what's d^2x/dt^2 of x=Asinωt?

Answer 4

you need 2 differentiations of x=Asinωt

Answer 5

I know I need to find the second derivative of the three equations too see that all of them are correct solutions. The problem is that I don't know which is the derivative of sine or cosine when t is multiplied by omega. Is the first derivative of the first equation just:\[\frac{ d }{ dt }Asen (\omega t)=Acos(\omega t)\]?

Answer 6

d(Asinωt)/dt=ωΑcosωt

Answer 7

you just multiply by ω every time you differentiate even it is in cos even it is in sin

Answer 8

And If I derive that again the i get \[\frac{ d }{ dt }\omega Acos(\omega t)=-w ^{2}Asen(\omega t)\]?

Answer 9

exactly ...you solved the 1st

Answer 10

Thanks! Now if I plug that into the solution I get:\[Asen(\omega t)=x\]And thats the same as \[Asen(\omega t +C)=x\]when I start at the point of equilibrium and therefore C (the constant of integration is zero). Right?

Answer 11

you are right. keep in mind that your C is φ that is the initial phase angle of simple harmonic motion

Answer 12

Thanks and how does the integration work in the third equation were I have and theta in the cos as well?

Answer 13

derivation sorry.

Answer 14

d(Acos(ωt+θ))/dt =ωΑsin(ωt+θ)

Answer 15

but that you need to understand why

Answer 16

because d(sin(ωt+θ))/dt=d(ωt+θ)cos(ωt+θ)

Answer 17

sorry d(sin(ωt+θ))/dt=d(ωt+θ)/dtcos(ωt+θ)

Answer 18

Well I guess thats were my calculus ability ends. I dont understand how there can be a function on the dt side. But don't worry I haven't seen calculus in highschool yet, I'm just trying to get a little bit ahead of mysself in physics. but thank you very much. Or do you think I can get it with only basic calculus knowledge (I know till implicit differeciation)

Answer 19

general rule---->(sin(ω)'=ω'cos(ω)

Answer 20

and cos(ω)'=ω'(-sin(ω))

Answer 21

Thanks

Answer 22

np..have fun with simple harmonic motion...one of the basic equations of physics:P