Solving Simple Harmonic Motion: Finding First Times for X=0 and V=0

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Homework Help Overview

The discussion revolves around solving a simple harmonic motion problem, specifically finding the first times when the position \( x \) and velocity \( v \) are both zero. The original poster provides initial conditions and an equation representing the motion, but encounters discrepancies between graphical and analytical results.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's attempts to solve the equation analytically, noting confusion over the correct formulation of the equation. There are inquiries about the use of trigonometric identities and the implications of periodicity in the motion.

Discussion Status

Some participants have offered guidance on using trigonometric identities to simplify the problem, while others have raised questions about the implications of finding multiple solutions for subsequent times when \( x = 0 \) and \( v = 0 \). The conversation reflects a mix of interpretations and approaches without reaching a consensus.

Contextual Notes

The original poster expresses uncertainty about the correct setup of the equation and the implications of periodic motion, indicating a need for clarification on these concepts. There is also mention of potential language barriers affecting understanding.

esradw
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I am trying to solve the part c of the question. I know the initial condidtions of the motion which are Xo=3 and Vo=50 my W=10 , B1=3 B2=5 and my equation is
x(t)=3cos(10t)+5sin(10t), the question is at which first times x=0 and x(2dot)=V=0,
Now, when I plot the equation on excel, I clearly see that at Xmax (Amax) V=0 at t=0.1 and when V=Vmax x=0 at around t=0.26s.

But when I want to find these times by solving the equation , I find something different.
how to solve this equation to find the first times at which x=0 and v=0, what am I missing while solving it ?

anyhelp appreciated

thanks
 
Last edited:
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esradw said:
I am trying to solve the part c of the question. I know the initial condidtions of the motion which are Xo=3 and Vo=50 my W=10 , B1=3 B2=5 and my equation is
x(t)=3cos(10t)=5sin(10t), the question is at which first times x=0 and x(2dot)=V=0,
Now, when I plot the equation on excel, I clearly see that at Xmax (Amax) V=0 at t=0.1 and when V=Vmax x=0 at around t=0.26s.

But when I want to find these times by solving the equation , I find something different.
how to solve this equation to find the first times at which x=0 and v=0, what am I missing while solving it ?

anyhelp appreciated

thanks

What do you mean by:
x(t)=3cos(10t)=5sin(10t)
? Are you actually subtracting the two?

Solving trig equations can be a real pain. There are a few guidelines, though, that can make it a bit easier.
1) Try to get all of the trig functions to the same argument. That's already done here.
2) Try to get all the trig functions to a single trig function. (This is the hard one...more below.)
3) Recall the limits of the trig functions, as they might help you decide where to focus your attention for zeros.
4) Have enough sanity to realize when and if it won't work and do it numerically! :)

Not much more to advise. I don't know of any hard and fast rules for solving them.

4) is not needed in this case!

Now, about 2). What I'm talking about here are trig identities. A few of them come up more often than not:
[tex]sin^2 \theta + cos^2 \theta = 1[/tex]
[tex]cos(2 \theta)=cos^2 \theta - sin^2 \theta = 2cos^2 \theta -1=1-2sin^2 \theta[/tex]
[tex]sin(2 \theta)=2sin \theta cos\theta[/tex]
The half-angle formulas are derived from these, but I haven't had many occasions to use them.

In your case the first relation looks good. Solve it for the cosine and plug it into your equation. That will give you a second order equation for cosine. Ugly, but it'll work.

-Dan
 
sorry ! I was spossed to write (+) instead of (=)

I tried to get squared of (X=...) but since I have coefficients of cos and sine so I can't use the (sin^2 \theta + cos^2 \theta = 1)

what do you mean with the first one can you explain or give an example ( Sorry English is my second language and sometimes I am having trouble with understanding)

thanks
 
Last edited:
esradw said:
sorry ! I was spossed to write (+) instead of (=)

I tried to get squared of (X=...) but since I have coefficients of cos and sine so I can't use the (sin^2 \theta + cos^2 \theta = 1)

what do you mean with the first one can you explain or give an example ( Sorry English is my second language and sometimes I am having trouble with understanding)

thanks

Try solving [tex]sin^2 \theta + cos^2 \theta = 1[/tex] for either sine or cosine. Then plug it into your equation.

-Dan
 
thanks ,Dan I will try it now :)

What I wonder is that , this is asking me the times at which X=0 and V=0 for the first time , what if It asked me the times at which X=0 and V=0 for the second time ? How do I solve that with my equation ? We know that the motion occurs every 2Pi/w time but how to use that to solve in this particular stuation.? Do you have any idea?
thanks again
 
esradw said:
thanks ,Dan I will try it now :)

What I wonder is that , this is asking me the times at which X=0 and V=0 for the first time , what if It asked me the times at which X=0 and V=0 for the second time ? How do I solve that with my equation ? We know that the motion occurs every 2Pi/w time but how to use that to solve in this particular stuation.? Do you have any idea?
thanks again

When you work out your solution it will be in terms of a sine or cosine function, depending on which you pick to work with. These functions are periodic so you will get a general solution sort of like [tex]t= \frac{2 \pi n}{3}[/tex] (I made that up) where n is some integer. By letting n be different integers, we get different points in the oscillating cycle.

-Dan
 

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