Simple Harmonic Problem: Finding Time at x=-0.05m with Mass on Spring

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In summary, the mass was pulled away from equilibrium and released at a point x=-.05m. The equation for calculating the time at which the mass is at that point was found using Acos(omega * t) and t was found to be .43 seconds.
  • #1
physics noob
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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

I am 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 can't seem to get the right answer.......any help is appriciated...THANKS
 
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  • #2
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.
 
  • #3
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?
 
  • #4
sorry to be a bother, but furhermore I am not getting .58 I am gettin nearer .42
 
  • #5
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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  • #6
youre the man...thanks a million!
 
  • #7
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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  • #8
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.
 
  • #9
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 don't see how else i can solve this problem
 
  • #10
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.
 
  • #11
physics noob said:
I am 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.
 
  • #12
ok i just got the right answer...by finding omega using f= omega/2pi

ok but now i have a couple of questions,,,,,, couldn't 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, doesn't that imply that omega is constant, meaning what would be the difference in omega at distance x=.18 doesn't velocity change through out the harmonic motion,,,,,,and if that's true angular velocity changes as well, becuase omega and V differ by either dividing or multiplying a constant R.

i guess I am 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
 
  • #13
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, I am pretty sure my professor said that's exactly what you can relate it too,,,, it has the same movement and velocity as circular motion
 
  • #14
physics noob said:
ok but now i have a couple of questions,,,,,, couldn't 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, doesn't 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 that's 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 I am 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.
 
  • #15
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!
 

Related to Simple Harmonic Problem: Finding Time at x=-0.05m with Mass on Spring

What is a simple harmonic problem?

A simple harmonic problem is a type of physical problem that involves a system that oscillates back and forth with a constant frequency. Examples include a pendulum or a mass on a spring.

What is the formula for calculating the period of a simple harmonic motion?

The formula for calculating the period of a simple harmonic motion is T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant.

How is a simple harmonic problem different from other types of physical problems?

A simple harmonic problem is different from other types of physical problems because it has a linear restoring force and a constant frequency. This makes it easier to analyze and solve mathematically.

What are the applications of simple harmonic motion in real life?

Simple harmonic motion is commonly observed in many real-life systems, such as pendulum clocks, musical instruments, and car suspensions. It is also used in fields such as engineering and physics to model and understand various systems.

What factors affect the frequency of a simple harmonic motion?

The frequency of a simple harmonic motion is affected by the mass of the object, the stiffness of the spring, and the amplitude of the oscillation. It is also independent of the initial conditions and the total energy of the system.

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