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

In summary, you should be able to integrate x = Acos(ωt+φ), where φ is an unknown phi, to find x(t), v(t), and a(t).
  • #1
Matt1234
142
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion in which a body oscillates back and forth around an equilibrium point with a constant period. It occurs when the restoring force on the body is directly proportional to its displacement from the equilibrium point and acts in the opposite direction.

What is the equation for Simple Harmonic Motion?

The equation for Simple Harmonic Motion is x(t) = A*sin(ωt + φ), where x(t) is the displacement of the body at time t, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

How do you find the velocity as a function of position in Simple Harmonic Motion?

The equation for velocity in Simple Harmonic Motion is v(t) = A*ω*cos(ωt + φ). To find the velocity as a function of position, we can use the equation v(x) = ±√(2E/m - 2kx²), where E is the total energy of the body, m is its mass, and k is the spring constant.

What is the relationship between velocity and acceleration in Simple Harmonic Motion?

In Simple Harmonic Motion, the acceleration of the body is directly proportional to its displacement from the equilibrium point and acts in the opposite direction. This means that the acceleration and velocity are out of phase by 90 degrees.

How does the mass and spring constant affect Simple Harmonic Motion?

The mass and spring constant affect the period and frequency of Simple Harmonic Motion. A higher mass will result in a longer period, while a higher spring constant will result in a shorter period. Additionally, the amplitude of the motion will decrease as the mass or spring constant increases.

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