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

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The discussion revolves around finding velocity as a function of position in simple harmonic motion, specifically addressing the integration challenges posed by the time variable in the equation. The initial approach involves using the equation x = A*cos(ωt + φ) to determine the phase constant φ, with the assumption that x = 0 corresponds to t = 0. Participants suggest integrating the acceleration equation a = -ω²x and provide a method to derive velocity through integration. The importance of accurately determining φ is emphasized, as it is crucial for solving subsequent parts of the problem. The conversation concludes with hints for calculating position and velocity at various time points.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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