Simple harmonic motion problem, finding velocity as a function of position

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Homework Help Overview

The discussion revolves around a simple harmonic motion problem, specifically focusing on finding velocity as a function of position. The original poster is attempting to determine the phase constant, phi, given the velocity and acceleration at a specific position.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between position, velocity, and acceleration in simple harmonic motion. There are attempts to integrate expressions related to velocity and acceleration, with some questioning the assumptions made about the initial conditions, particularly regarding the timing when position is zero.

Discussion Status

Some participants have offered guidance on how to approach the integration and the determination of phi. There is an ongoing exploration of the implications of the assumptions made about the initial conditions, with no explicit consensus reached on the correct interpretation.

Contextual Notes

There is a noted lack of specification in the problem regarding the relationship between position and time at the moment when position is zero, leading to questions about the validity of certain assumptions.

Matt1234
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Homework Statement


2lv1v9u.png


I need help finding velocity as a function of position, the t in the argument of tan, causes a problem for me when integrating. can someone help, maybe my approach is completely off, i am trying to fing the unknown phi, knowing the velocity and acceletation when position is 0.

thanks for your time.
matt
 
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So you have x= Acos(ωt+φ), you should know that when x=0,t=0 meaning that you can get the value of φ easily.

As for the integration, once you get φ, it would be easier to start with

a=-ω2x

v(dv/dx)=-ω2x

v dv = -ω2x dx

∫v dv = -ω2∫ x dx.
 
rock.freak667 said:
So you have x= Acos(ωt+φ), you should know that when x=0,t=0 meaning that you can get the value of φ easily.

As for the integration, once you get φ, it would be easier to start with

a=-ω2x

v(dv/dx)=-ω2x

v dv = -ω2x dx

∫v dv = -ω2∫ x dx.

the problem doesn't specify that when x = 0, t =0, how can i make such an assumption?
 
Matt1234 said:
the problem doesn't specify that when x = 0, t =0, how can i make such an assumption?

Yes you are right, I thought your expression for x had the sine term in it, sorry about that.
 
no problem, i think this boils down to integrating that above statement, I am sorry for the poor image quality, my scanner broke.
somehow i need phi, I am not quite sure how to obtain it. i think parts a and b we intentional to get me thinking in the way of integration.
 
Matt1234: You can see, at t = t1 = 0, x1 = 0.10 m. In other words, you need to fully deflect the mass, then release it, in order for it to vibrate. Therefore, at t1 = 0, we have,

x1 = 0.10 m = (0.10 m)*cos(4*t1 + phi)
cos(4*t1 + phi) = 1
4*t1 + phi = acos(1)
phi = acos(1) - 4*t1
phi = 0 - 4*0
phi = 0 rad​

Therefore, we have,

x(t) = A*cos(omega*t + phi)
x(t) = A*cos(omega*t + 0)
x(t) = A*cos(omega*t)​

Likewise, v(t) = -A*omega*sin(omega*t).

(c) Let x(t2) = 0.06 m. Therefore,

x(t2) = A*cos(omega*t2)
0.06 = 0.10*cos(4*t2)​

Hint 1: Can you solve for t2? After that, can you compute v(t2)?

(d) Hint 2: Can you compute a(t2)?

(e) Hint 3: Let x(t3) = 0 m. Can you solve for t3? Hint 4: Let x(t4) = -0.08 m. Can you solve for t4?

Try again. Also, please do not post wide images directly to the forum page. Just post a text link to wide images.
 

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