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

In summary, the conversation discusses solving a physics problem involving initial conditions of motion and a trigonometric equation. The goal is to find the first times at which x=0 and v=0, but there is confusion on how to solve the equation to get these times. The expert advises using trig identities and solving for either sine or cosine to get a second order equation, which will give a general solution in terms of n, an integer representing different points in the oscillating cycle. The expert also mentions that the motion occurs every 2Pi/w time, but it is unclear how to use this information in this particular situation.
  • #1
esradw
26
0
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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  • #2
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
 
  • #3
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:
  • #4
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
 
  • #5
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
 
  • #6
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
 

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

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement of the object from its equilibrium position. This means that the object oscillates back and forth around its equilibrium point, with a constant period and amplitude.

2. What are the key characteristics of simple harmonic motion?

The key characteristics of simple harmonic motion are its period, amplitude, and frequency. The period is the time it takes for one complete oscillation, the amplitude is the maximum displacement of the object from its equilibrium position, and the frequency is the number of oscillations per unit time.

3. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A sin(ωt + φ), where x is the displacement of the object from its equilibrium position, A is the amplitude, ω is the angular frequency, and φ is the phase angle. This equation can also be written as x = A cos(ωt + φ).

4. What is the relationship between simple harmonic motion and energy?

Simple harmonic motion is a conservative system, meaning that energy is conserved within the system. As the object oscillates back and forth, its potential energy and kinetic energy are constantly changing, but the total energy remains constant. At the equilibrium position, all of the energy is potential energy, and at the maximum displacement, all of the energy is kinetic energy.

5. What are some real-life examples of simple harmonic motion?

Some real-life examples of simple harmonic motion include a pendulum, a mass on a spring, and a mass attached to a vertical spring. These systems exhibit simple harmonic motion due to the force of gravity or the force of the spring being directly proportional to the displacement of the object. Other examples include a guitar string, a swing, and a vibrating tuning fork.

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