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Yet another simple harmonic problem

yet another "simple" harmonic problem

1.01kg mass is attached to a horizontal massless spring with k=10N/m The system is set into motion by pulling the mass .2m away from equilibrium, releasing it from rest.

find a time when the mass is at x= -.05m

so i used x=Acos(omega *t)

so to get omega i used MEi = MEf
getting .5(10)(.2)^2 = .5(10)(.05)^2 + .5(1.01)v^2

im gettin a velocity of .61 which is an angular velocity of

3.05 rad/s i plug it into the aforemention equation

so .05=.20cos(3.05(t)) t = .43 sec which of course is wrong correct answer is .58

where did i go wrong here,,,,, also i have tried several combinations starting Xo at different spots. but i cant seem to get the right answer.............................any help is appriciated.........THANKS
 

Pyrrhus

Homework Helper
2,160
1
This is a simple plug in the formula...

[tex] \omega = \sqrt{\frac{k}{m}} [/tex]

[tex] A = 0.2 m [/tex]

[tex] x = A \cos (\sqrt{\frac{k}{m}} t) [/tex]

Simply solve for t.
 
hmmm. it would have been an easy plug in the answers if i had known that

omega = sqroot of k/m is this commonly known? because my professor never went over that equation.... can you derive it?
 
sorry to be a bother, but furhermore im not getting .58 im gettin nearer .42
 

Pyrrhus

Homework Helper
2,160
1
Yes i can derive it.

Now let's study the simple harmonic motion of a spring-mass system, by using Newton's 2nd Law in scalar form:

[tex] \sum_{i=1}^{n} F_{i} = ma [/tex]

There's a restorative force spring force

[tex] -kx = ma [/tex]

which is a 2nd order ODE

[tex] \frac{k}{m}x + \ddot{x} = 0 [/tex]

Looking at the standard Simple Harmonic motion equation

[tex] \omega^{2} x + \ddot{x} = 0 [/tex]

we get

[tex] \omega = \sqrt{\frac{k}{m}} [/tex]

Remember to put your calculator in radians.
 
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youre the man...............thanks a million!!!
 

OlderDan

Science Advisor
Homework Helper
3,021
1
The angular velocity (more appropriately called angular frequency in this problem) has nothing to do with how much energy you give the system. It depends on mass and spring constant. Find the appropriate relationship and calculate it.

Don't forget the minus sign when you go to solve for the time.
 
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Doc Al

Mentor
44,718
1,020
physics noob said:
PS i posted this yesterday, and despite some helpful responses i have yet to obtain the correct answer
Please do not start a second thread with the same problem! I merged OlderDan's post into this existing thread.
 
The angular velocity (more appropriately called angular frequency in this problem) has nothing to do with how much energy you give the system.




yes but velocity does, which i then used to find angular velocity, which i then used to find time......i dont see how else i can solve this problem
 

Doc Al

Mentor
44,718
1,020
physics noob said:
1.01kg mass is attached to a horizontal massless spring with k=10N/m The system is set into motion by pulling the mass .2m away from equilibrium, releasing it from rest.

find a time when the mass is at x= -.05m

so i used x=Acos(omega *t)
That will work. First find omega from the mass and spring constant. You have x and A.

As OlderDan says, omega has nothing to do with the energy given to the system, so no need for conservation of energy.
 

Doc Al

Mentor
44,718
1,020
physics noob said:
im gettin a velocity of .61 which is an angular velocity of

3.05 rad/s
You are confusing simple harmonic motion, which has an angular frequency, with circular motion which has an angular speed. [itex]v = \omega r[/itex] is not relevant here.
 
ok i just got the right answer......by finding omega using f= omega/2pi

ok but now i have a couple of questions,,,,,, couldnt i find omega by finding the velocity when the PE of the spring is .05......by using MEi=MEf....then when i have velocity i can convert it to angular velocity by dividing by .2 which is the radius........and then put that into x=Acos(omegaT) and when i found omega using the above equation, doesnt that imply that omega is constant, meaning what would be the difference in omega at distance x=.18 doesnt velocity change through out the harmonic motion,,,,,,and if thats true angular velocity changes as well, becuase omega and V differ by either dividing or multiplying a constant R.

i guess im most confused on how omega is constant through out this process of harmonic motion , when acceleration and velocity are constantly changing.

oh thanks are in order for all of you who helped//////thanks again
 
Doc Al said:
You are confusing simple harmonic motion, which has an angular frequency, with circular motion which has an angular speed. [itex]v = \omega r[/itex] is not relevant here.

wait, im pretty sure my professor said thats exactly what you can relate it too,,,, it has the same movement and velocity as circular motion
 

Doc Al

Mentor
44,718
1,020
physics noob said:
ok but now i have a couple of questions,,,,,, couldnt i find omega by finding the velocity when the PE of the spring is .05......by using MEi=MEf....then when i have velocity i can convert it to angular velocity by dividing by .2 which is the radius........and then put that into x=Acos(omegaT)
No. As I said earlier, do not confuse uniform circular motion (where [itex]v = \omega r[/itex]) with simple harmonic motion.
and when i found omega using the above equation, doesnt that imply that omega is constant, meaning what would be the difference in omega at distance x=.18
Omega is constant!
doesnt velocity change through out the harmonic motion,,,,,,and if thats true angular velocity changes as well, becuase omega and V differ by either dividing or multiplying a constant R.
Linear speed does change, but the angular frequency does not.

i guess im most confused on how omega is constant through out this process of harmonic motion , when acceleration and velocity are constantly changing.
You can relate uniform circular motion to SHM like this: Imagine something moving in a circle at a uniform angular rate [itex]\omega[/itex]. The x-component of that circular motion executes SHM (described by [itex]A\cos \omega t[/itex]). The speed of that x-component varies from 0 to a maximum of [itex]A \omega[/itex], but [itex]\omega[/itex] remains constant.
 
ahhh, i see now,,,, cause in the circular motion only the x velocity is pertantent to SHM......thanks a million doc,,,, you run a great place here!!!!
 

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