Solved: Harmonic Motion - A, T, v=0, Acceleration, & Phase

[BlueRiboon]

Homework Statement

The motion of a particle is given by: x = (6.0m) cos(0.586t + 0.72)

a) Find the amplitude
b) Find the Period
c) Find the first time for t>0 when v=0
d) Find the maximum acceleration
e) Fine the phase at time t=1.38s

Homework Equations

x = A cos ωt
T = 2pi/ω
f = 1/T

The Attempt at a Solution

I am taking the course on Online, there is no online lecture..
Have to learn everything from the textbook but, English is my second language :'(
Having hard time understand the basic concept..

a) A = 6.0

b) I don't know about 0.586t + 0.72
should I just calculate like.. T = 2pi / ω , T =2pi/0.586 ?
Where do I use 0.72?

c) v=0 (t>0) it means, fine the time at T/2 ?

d) no idea..

e) is phase means to find x ? so just plug t=1.38 in the formula?


Thank you so much for your help.. quicker the better..

I am not even done with studying everything , but have a final exam in 2 days.
.. I started the course 5 weeks ago with another offline course.. (due to school's cancellation of my course) , and so far finished 12 chapters.., 2 more chapters to go!

Any help would be so appreciated!
Thanks!

 

[lewando]

Try exploring this equation by plotting it [fooplot.com provides a nice tool].
For d), you are given x(t). v(t) = dx(t)/dt, a(t) = dv(t)/dt.

 

[BlueRiboon]

Try exploring this equation by plotting it [fooplot.com provides a nice tool].
For d), you are given x(t). v(t) = dx(t)/dt, a(t) = dv(t)/dt.

Thank you for your reply

I tried plotting it that website,

For a) b) d) , After I plotting on the website, I think +0.72 just shift whole thing to the right,
so won't effect the amplitude, period and max Accelation.

If my assumption is correct, I got the a,b,d but still have no idea about c)
and, for e) do I just have to put time in the equation?

Thanks!

 

[PeterO]

Homework Statement

The motion of a particle is given by: x = (6.0m) cos(0.586t + 0.72)

a) Find the amplitude
b) Find the Period
c) Find the first time for t>0 when v=0
d) Find the maximum acceleration
e) Fine the phase at time t=1.38s

Homework Equations

x = A cos ωt
T = 2pi/ω
f = 1/T

The Attempt at a Solution

I am taking the course on Online, there is no online lecture..
Have to learn everything from the textbook but, English is my second language :'(
Having hard time understand the basic concept..

a) A = 6.0

b) I don't know about 0.586t + 0.72
should I just calculate like.. T = 2pi / ω , T =2pi/0.586 ?
Where do I use 0.72?

c) v=0 (t>0) it means, fine the time at T/2 ?

d) no idea..

e) is phase means to find x ? so just plug t=1.38 in the formula?


Thank you so much for your help.. quicker the better..

I am not even done with studying everything , but have a final exam in 2 days.
.. I started the course 5 weeks ago with another offline course.. (due to school's cancellation of my course) , and so far finished 12 chapters.., 2 more chapters to go!

Any help would be so appreciated!
Thanks!

b) If a weighted spring was set in oscillation in shm, and you then walked into the room and watched it - what is to say it was passing through the equilibrium position at the instant you walked in? The 0.72 is the factor necessary for the mathematical description to match what you actually observed - so you proposal for calculating T looks good.

c) v=0 occurs at maximum displacement - so when cos(...) = 1 and -1. What is the smallest positive value of t that meets those conditions.

d) maximum acceleration occurs at maximum displacement, though using calculus [derivatives] as suggested by lewando is probably called for.

e) not quite sure what is meant by this.

 

[Nickie]

Hello, it seems like you are struggling with some of the concepts in harmonic motion. Let's go through each question together to help you understand the concepts better.

a) The amplitude is the maximum displacement from equilibrium, in this case it is 6.0m. So you are correct, the amplitude is 6.0m.

b) The period is the time it takes for one full cycle of the motion. In this case, the period can be found using the formula T = 2π/ω, where ω is the angular frequency. In your equation, ω = 0.586, so you can plug that into the formula to find the period. The value of 0.72 is the phase angle, which we will discuss in question e).

c) To find the first time when v=0, we need to look at the velocity equation, which is v = -ωA sin(ωt + Φ), where Φ is the phase angle. In this case, we can see that when the sine term is equal to 0, the velocity will be 0. So, we can set ωt + Φ = 0 and solve for t. This will give us the first time when v=0.

d) The maximum acceleration occurs at the equilibrium point, where x=0. Therefore, we can find the maximum acceleration by plugging in x=0 into the acceleration equation, which is a = -ω^2A cos(ωt + Φ). This will give us the maximum acceleration value.

e) The phase angle, Φ, represents the initial position of the particle at t=0. In this case, the phase angle is 0.72. So, to find the phase at t=1.38s, we can plug in t=1.38 into the equation, giving us x = (6.0m) cos(0.586(1.38) + 0.72). This will give us the position of the particle at t=1.38s, but it is not the phase angle.

I hope this helps you understand the concepts better. Remember to study and practice as much as you can before your exam. Good luck!