How Is Kinetic Energy Related to Spring Compression in Harmonic Motion?

[artsakh]

A 4 kg mass is attached to a 100 N/m spring and oscillates with an amplitude of 0.75 m
across a horizontal frictionless surface. When the kinetic energy is 70% of the total
mechanical energy, by how much is the spring stretched or compressed?

I'm not sure how to approach this problem. I could figure out the \omega by sqrt(k/m) and then use it in the x=acos(\omegat+\phi) formula, but what do i do with the 70%? Multiply the amplitude by .70??

A spring has a spring constant 120 N/m. If a 10 kg mass is attached to it and undergoes
simple harmonic motion, how long does it take for the mass to move from one end (when the
spring is most stretched) to the other (when the spring is most compressed)?

I used the period formula: 2pi/\omega, \omega=sqrt(k/m), (2pi/sqrt(k/m))/2 since the period is back and forth. I got the right answer, but I am wondering, was it by luck? Is the reasoning correct? If not, what did i do wrong?

Any help would be greatly appreciated as I am studying for a final. Thank you very much.

 

[Rayquesto]

I believe that when kinetic energy exists 70 percent of mechanical energy there exists potential energy that makes up mechanical energy. so, I believe that .7KE=.3PE I think

 

[Rayquesto]

so 4 times 100 times x^2 times .3= 1/2 times 4 times a velocity .7

to get velocity you must use v=frequency times the wavlength which is .75meters

why don't we just assume that frequency is 1, because of the behavior of the wave.

so the velocity of the wave is .75m/s

120x^2= 1.4 times .75
and you do the math
dude I could be wrong when I set up velocity
check in with someone else as to how to find velocity or rather how to find frequency

 

[Rayquesto]

The period formula works for pendulums 20 degrees and under in terms of the angle of reflection and index. By memory, I think the formula is T-2pie times the square root of Length divided by gravitational force.

In terms of a spring the period formula works all the time (I think): 2 pie times the square root of Mass divided by spring constant.

 

[Rayquesto]

so 4 times 100 times x^2 times .3= 1/2 times 4 times a velocity .7

to get velocity you must use v=frequency times the wavlength which is .75meters

why don't we just assume that frequency is 1, because of the behavior of the wave.

so the velocity of the wave is .75m/s

120x^2= 1.4 times .75
and you do the math
dude I could be wrong when I set up velocity
check in with someone else as to how to find velocity or rather how to find frequency

Ok correction: frequency can be found using knwledge of period. here's how to do it: use 2 pie times the sqrt of m/k to find period then take inverse. ok so the mass is 4 kg and the constant is 100. 4/100 is .04 and sqrt of .04 is .2 and multiply that times 2 to get .4 and multiply that times pie to get 1.256 seconds. take the inverse of that to get .80 s^-1

now take .8 times .75 and you have yourself a velocity of .6m/s

so now its actually
120x^2=1.4 times .6m/s
x^2=.84/120
x=sqrt.007 lol .007 james bond
to get .084meters stretched lol that seem a bit small don't you think?

 

[Artius]

I hope someone answers this before 12:30 EST

 

[lepton5]

A 4 kg mass is attached to a 100 N/m spring and oscillates with an amplitude of 0.75 m
across a horizontal frictionless surface. When the kinetic energy is 70% of the total
mechanical energy, by how much is the spring stretched or compressed?

I'm not sure how to approach this problem. I could figure out the \omega by sqrt(k/m) and then use it in the x=acos(\omegat+\phi) formula, but what do i do with the 70%? Multiply the amplitude by .70??

A spring has a spring constant 120 N/m. If a 10 kg mass is attached to it and undergoes
simple harmonic motion, how long does it take for the mass to move from one end (when the
spring is most stretched) to the other (when the spring is most compressed)?

I used the period formula: 2pi/\omega, \omega=sqrt(k/m), (2pi/sqrt(k/m))/2 since the period is back and forth. I got the right answer, but I am wondering, was it by luck? Is the reasoning correct? If not, what did i do wrong?

Any help would be greatly appreciated as I am studying for a final. Thank you very much.

just straight to the problem:
1) you know that EM = \frac{1}{2} k A^2 for conservation of EM = EP + EK.
When EK = 70% EM, then the eqn become EM = EP + 0.7 EM, if EP = \frac{1}{2} k x^2 then you can calculate it to solve for x! you will get the value for x about 0.411 m.

2) I think your start is correct with the angular freq, since from the definition of T (period time) that is the time needed for one complete oscillation. (I guess you should divide by 2 from T, since the question ask for the time when spring most stretch to most compressed).

hope it can help and not late...

 

[artsakh]

just straight to the problem:
1) you know that EM = \frac{1}{2} k A^2 for conservation of EM = EP + EK.
When EK = 70% EM, then the eqn become EM = EP + 0.7 EM, if EP = \frac{1}{2} k x^2 then you can calculate it to solve for x! you will get the value for x about 0.411 m.

2) I think your start is correct with the angular freq, since from the definition of T (period time) that is the time needed for one complete oscillation. (I guess you should divide by 2 from T, since the question ask for the time when spring most stretch to most compressed).

hope it can help and not late...

Thank you very much! Trying to get all this in before my final tomorrow, thanks a lot.

 

[Rayquesto]

dang I am sorry i totally calculated stuff incorrectly. if 70 percent of total mechanical energy is made up of kinetic energy, then that doesn't mean that 30 percent is mad up of potential energy. I am really sorry. so this other guy was right. and if total mechanical energy is constantly KE plus PE, then its easy to solve. you just say .7ME=KE and ME=KE plus PE, and then you replace the first equation by saying ME=.7ME plus PE and yea you get .411 meters streched. haha this other guy is a genius! I'll try to help you with the other problem. hold on.

 

[Rayquesto]

A spring has a spring constant 120 N/m. If a 10 kg mass is attached to it and undergoes
simple harmonic motion, how long does it take for the mass to move from one end (when the
spring is most stretched) to the other (when the spring is most compressed)?

for the second problem, you could first find its maximum stretch distance meaning when mechanical energy is only potential energy, but did your professor or your mind tell you that in a spring, the times when it is most stretched and most compressed are when potential energy is at its highest and so is acceleration?

So, I'm pretty sure you just find period where T= 2 pie times sqrt mass divided by constant, since period is defined as time going from one point of max potential energy to the next.

So, T=2pie times sqrt of 10/120=1.82 seconds

 

[artsakh]

A spring has a spring constant 120 N/m. If a 10 kg mass is attached to it and undergoes
simple harmonic motion, how long does it take for the mass to move from one end (when the
spring is most stretched) to the other (when the spring is most compressed)?

for the second problem, you could first find its maximum stretch distance meaning when mechanical energy is only potential energy, but did your professor or your mind tell you that in a spring, the times when it is most stretched and most compressed are when potential energy is at its highest and so is acceleration?

So, I'm pretty sure you just find period where T= 2 pie times sqrt mass divided by constant, since period is defined as time going from one point of max potential energy to the next.

So, T=2pie times sqrt of 10/120=1.82 seconds

You're right but you have to divide that number in half