Oscillatory Motion - Determining equation of motion

[menglish20]

Homework Statement

A particle with a mass of 0.5 kg is attached to a horizontal spring with a force constant of 50 N/m. At the moment t = 0, the particle has a maximum speed of 20 m/s and is moving to the left.
(a) Determine the particle's equation of motion.
(b) Where in the motion is the potential energy three times the kinetic energy?
(c) Find the minimum time interval required for the particle to move from x=0 to x=1.00m.
(d) Find the length of a simple pendulum with the same period.

Homework Equations

(a)
w=sqrt(k/m)

v_max = Aw

(b)
3(.5*m*w^2*A^2*sin^2(wt+phi))= .5*k*A^2*cos^2(wt+phi)

The Attempt at a Solution

I searched he forums here for the same question and found this thread:
https://www.physicsforums.com/archive/index.php/t-231270.html

I think I understand how to find omega and the maximum velocity (though the signs may be incorrect), but I don't understand how to solve for phi. The only thing I could think of was to set

v=-wAsin(wt+phi) to -20=10*(-2)sin(phi) for t=0.
This returned a phi=3pi/2 or -pi/2, unless I'm doing something wrong.
I also tried to set x(0)=0, so
0=(-2)cos(phi) which results in phi=pi/2.
These answers disagree with what the linked thread found for phi and what my professor's answer had for phi.

I'm just having a hard time understand phi, which seems like such a simple concept. If anyone could steer me in the right direction, I think I could finish this problem. Thanks for any help!

 

[ideasrule]

v=-wAsin(wt+phi) to -20=10*(-2)sin(phi) for t=0.

If -20=10*(-2)sin(phi), then sin(phi)=1, so phi can be pi/2. This is consistent with what you got using x(0)=0.

 

[gneill]

Just a thought: if the speed is maximum as the particle passes through the equilibrium point at t=0, why not make the velocity function a cosine and do away with the phase constant? Integrate once to find the position function (which will then be a sine function, again with no phase constant). Should make life easier.

 

[bgc]

If you're going to do more with oscillators (or waves) I recommend purchasing A. P. French's Vibrations and Waves It's inexpensive -- likely free down-loadable -- there are MIT lectures based on it.

bc

 

[rwisz]

Firstly, it's great that you have attempted to solve the problem by looking at previous threads and using relevant equations. As a scientist, it's important to seek out and understand different approaches to solving problems.

Now, let's address your questions. In this case, the equation of motion for the particle can be written as x(t) = A*cos(wt + phi), where A is the amplitude (maximum displacement from equilibrium) and phi is the phase constant. The maximum velocity of the particle is given by v_max = A*w*sin(wt + phi).

To find the value of phi, you can use the initial conditions given in the problem. At t=0, the particle has a maximum speed of 20 m/s and is moving to the left. This means that v_max = -20 m/s. Substituting this into the equation for v_max, we get -20 = A*w*sin(phi).

Now, we can use the value of A and w (which you have already correctly calculated) to solve for phi. This gives us phi = -pi/2 or 3pi/2. Both solutions are correct, as they correspond to the particle moving to the left with a maximum speed of 20 m/s. The difference in the signs of phi is simply due to the different conventions used for the direction of rotation in trigonometric functions.

For part (b), you can use the equations for potential and kinetic energy (PE = 0.5*k*A^2*sin^2(wt + phi) and KE = 0.5*m*w^2*A^2*cos^2(wt + phi)) to solve for the time when the potential energy is three times the kinetic energy. This will give you an equation involving phi, which you can then solve for the corresponding value of t.

For part (c), you can use the equation x(t) = A*cos(wt + phi) and the given initial and final positions to solve for the time interval t.

For part (d), you can use the equation for the period of a simple pendulum (T = 2*pi*sqrt(L/g)) and equate it to the period of the oscillating particle (T = 2*pi/w). Solve for the length L to get the answer.

I hope this helps and clarifies your understanding of the problem. Keep up the good work in seeking out different approaches and asking for help when