Finding Amplitude for Horizontal Oscillation - 200 g Mass, 2.0 Hz

[nicksim117]

Hello, I am having trouble finding the A in this problem. I found T=1/2=.5, I found K =31.55n/m. I tried using the formula KA2=MV2+KX2 to find amplitude but ended up with A=0.0435m
This just doesn't seem right . I used .05m for a value of X. Does anybody know what I am doing wrong?



A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At one instant, the mass is at x=5.0cm and has Vx=.30 cm/s determine the period, Vmax, Amplitude, and the total energy.

 

[nicksim117]

The V should be Vx=-30cm/s

 

[alphysicist]

Hi nicksim117,

Can you list the details of what you did to get 0.0435m? It looks to me like you have a calculation error (or maybe you're interpreting the equation incorrectly?). Did you square the velocity?

 

[Redbelly98]

Everything in the equation is a positive quantity, since v and x get squared. There is just no way for A to end up being less than x.

 

[RTW69]

k looks ok

use energy methods E=K+U

you know v,x and k solve for A. the rest is pretty straight forward.

 

[nicksim117]

Homework Statement

Hello, I am having trouble solving this problem trying to find A I keep getting an amplitude which is less than my distance can anybody see what i am doing wrong?

A 200g mass attached to a horizontal spring oscillates at a frequency of 2.0 hz. At one instant, the mass is at X=5cm and has a V=-30cm/s Determine:
The period T=.5
The amplitude?
the Vmax
the total energy

Homework Equations

KA2=MV2+KX2 or A2=(MV2+KX2)/K

The Attempt at a Solution

solve for A using 1/2KA2=1/2MV2+1/2KX2= (.5)(31.55)A2=
> (.5)(.2)(-.3m/s)+(.5)(31.55)(.05m)2
> ==15.77A2=-.009+.0394375==.0304375/15.77=A2===.001930==A=.04393m
> V=-30cm/s
> K=31.55n/m
> X=5cm
> M=.2kg

A=.04393m

 

[nicksim117]

I don't know why i keep getting that answer. I posted the equation worked out as a new thread.

[Edit by Doc Al: I merged the two threads. Do not create multiple threads on the same problem.]​

 

[Doc Al]

solve for A using 1/2KA2=1/2MV2+1/2KX2= (.5)(31.55)A2=
> (.5)(.2)(-.3m/s)+(.5)(31.55)(.05m)2
> ==15.77A2=-.009+.0394375==.0304375/15.77=A2===.001930==A=.04393m

Your approach is fine, but recheck your arithmetic.

 

[alphysicist]

The Attempt at a Solution

solve for A using 1/2KA2=1/2MV2+1/2KX2= (.5)(31.55)A2=
> (.5)(.2)(-.3m/s)+(.5)(31.55)(.05m)2
> ==15.77A2=-.009+.0394375==.0304375/15.77=A2===.001930==A=.04393m
> V=-30cm/s
> K=31.55n/m
> X=5cm
> M=.2kg

A=.04393m

You squared the velocity, so it should be positive 0.009

 

[Doc Al]

You squared the velocity, so it should be positive 0.009

There you go!

 

[nicksim117]

Thank you!

 

[nicksim117]

For the same problem i am ask to report total energy in the system. When I use the formula and value at that instantE=K+U=1/2MV2+1/2KX2 I get .0484J. When I use the same formula but Vmax and A are used in place of V and X I get .0954J Should these values be the same because energy is conserved.

 

[Doc Al]

When I use the same formula but Vmax and A are used in place of V and X I get .0954J

Do v=Vmax and x=A occur at the same point?

Energy is most definitely conserved--that should tell you that something is wrong with your thinking.

 

[nicksim117]

Ok, that make sense, so If energy is conserved and I use the values given in the problem and plug them into E=K+U=1/2MV2+1/2KX2 to find total energy in the system they should equal the value of either Umax 1/2 KA2or Kmax 1/2MV2 not both added together.

 

[Doc Al]

Exactly.

 

[nicksim117]

thanks for your help!