Recent content by zachmgilbert

Thank You, that gave me the right answer.

i have .5mv2=.5kx2 but i don't know how to find velocity.

It is a giancoli problem so I can reset the variables. I've tried it multiples times and the F=-kx is always half of the correct answer. Where does it get multiplied by 2.

I tried F=-kx using mg for force. mg is (.25)(9.8) which equals 2.45. Dividing by 5.4 gave me .45 which is wrong. The correct answer is .91

Homework Statement A 0.25 kg mass is attached to a spring with spring constant 5.4 N/m and let fall. To the nearest hundredth of a meter what is the point where it 'stops'? diagram here: http://wps.prenhall.com/wps/media/objects/1088/1114633/ch11/grav.gif Homework Equations...

angular frequency squared

Homework Statement What is the potential energy of the spring? mass=.52 kg velocity=-6.54sin(20.1t) time=40.84 seconds amplitude=.33 meters angular frequency=20.1 Homework Equations U =1/2kx2 The Attempt at a Solution I used x=Acos(wt) and got .2 meters for x. I know this...

You are correct. I thought v=-6.54sin(20.1t) was a form of v=λf and I couldn't figure out where the sin and t came from. Thank you for helping me with a stupid mistake.

update: i used T=1/f to find time, divided by four because that's when the displacement will be maximum plugged all this in and got .32, the correct answer I know is .33. I didn't round to the end so I feel like this is luck and not the correct way, am i making this too difficult?

Homework Statement what is the amplitude of the motion v=-6.54sin(20.1t) (mks units) angular frequency=20.1 Homework Equations v=wAcoswt The Attempt at a Solution I think the wavelength is 6.54 since 20.1 is the frequency. v =6.2 m/s. I am missing time.