Angular frequency of a spring-mass system?

[MCATPhys]

Here's my question: say a car with x kinetic energy crashes into a spring so that the elastic potential energy becomes x. During the process, the spring compresses a distance of 9.13 meters. How do I find the time it takes for the spring to compress that distance (what information will I need?). The frequency is the number of oscillations per second, and the period is the number of seconds in one oscillation.

So... in my head... it feels like the time would be 1/4 the period - because that would give the time the mass reached the highest amplitude. Is this right? There's also something called angular frequency.. but I have no idea what that is.

 

[pgardn]

Here's my question: say a car with x kinetic energy crashes into a spring so that the elastic potential energy becomes x. During the process, the spring compresses a distance of 9.13 meters. How do I find the time it takes for the spring to compress that distance (what information will I need?). The frequency is the number of oscillations per second, and the period is the number of seconds in one oscillation.

So... in my head... it feels like the time would be 1/4 the period - because that would give the time the mass reached the highest amplitude. Is this right? There's also something called angular frequency.. but I have no idea what that is.

Do you have any equations that go with simple harmonic motion of a mass spring system?

 

[MCATPhys]

this question is an example of simple harmonic motion...

 

[pgardn]

this question is an example of simple harmonic motion...

Yes it is.

Do you have any equations that describe simple harmonic motion of a mass spring system? As a hint it looks like it requires conservation of energy. But the only info you have been given is the amount the spring compresses. Were you given anything else? Something about the spring's ability to stretch or compress? Or are you supposed to describe only using x and the info about the amount the spring compressed?

There are equations that desribe kinetic and elastic potential energy. And if the spring was really hard to compress the time it would take to turn all the kinetic energy into elastic pot. energy would be diff. The mass of the car would make a difference as well. Any equations?

 

[MCATPhys]

sorry... the spring is attached to the front of a car and it collides with a wall... and I'm asked to calculate the time it takes for the car to come to a full stop with the wall. the time it takes the car to reach the wall is 1 second. therefore, the total time is 1 sec + the time it takes for the spring to compress.

the spring constant is 3000, the mass is 100 kg, the velocity of the car is 50 m/s. I am also given the force exerted by the car on the wall, which is 25 * 10^6

the truth is I was actually hoping someone could explain the problem, instead of giving me a number.

the answer is 1/4 * square root of k/m = 1.3 seconds

 

[MCATPhys]

I added more details above

 

[pgardn]

sorry... the spring is attached to the front of a car and it collides with a wall... and I'm asked to calculate the time it takes for the car to come to a full stop with the wall. the time it takes the car to reach the wall is 1 second. therefore, the total time is 1 sec + the time it takes for the spring to compress.

the spring constant is 3000, the mass is 100 kg, the velocity of the car is 50 m/s. I am also given the force exerted by the car on the wall, which is 25 * 10^6

the truth is I was actually hoping someone could explain the problem, instead of giving me a number.

the answer is 1/4 * square root of k/m = 1.3 seconds

Ok it really does not make a difference if the spring is on the wall or the car.

So basically all the kinetic energy is converted to elastic potential energy in the time it takes for the car to go from 50 m/s to 0 m/s. But now that I see they have given you the force exerted by the car on the wall I can see using impulse and the change in momentum caused by the impulse. Are you familiar with this? Impulse and change in momentum? We can use this or the conservation of energy. But we must assume the force exerted by the wall on the car, or vice versa remains the same throughout the collision (the compression of the spring) Does it say anything about the Force remaining constant or that the force exerted is the average throughout the time?

 

[MCATPhys]

it doesn't really give that information... but the impulse idea actually makes sense.. but i get the wrong answer...

i don't know if this matters - but the way they showed the answer - it looks like it's 1/4th the angular frequency. the angular frequency is equal to square root (k/m). Do you know how the angular frequency relates to the compression time?

 

[pgardn]

it doesn't really give that information... but the impulse idea actually makes sense.. but i get the wrong answer...

i don't know if this matters - but the way they showed the answer - it looks like it's 1/4th the angular frequency. the angular frequency is equal to square root (k/m). Do you know how the angular frequency relates to the compression time?

Well the 1/4 probably comes from the following.

When the car with spring attached first strikes the wall and then comes to a halt, well this would be 1/4 of the entire period of this mass spring system. In other words if we kept watching what happens after the mass comes to rest, the spring would push the mass back in the other direction until it reached the point where the spring and the wall first came in contact, then it would go past this point backwards away from the wall (assume the spring is now stuck to the wall) until it came to rest, then it would move back towards the wall and reach the middle point again (equilibrium position) and that would be the entire trip. 1/4 of the period to go from 50 to zero, 1/4 of the period to get pushed back to the point where the spring and the wall first made contact and the car is going 50 m/s in the other direction, 1/4 of the time to go past this point and reach the point where the car was moving away from the wall (pretending like the spring is still stuck to the wall) and now had reached the other extreme going zero m/s, and then the last fourth of the trip would be on the way back to the very first position where the spring made contact with the wall. This would be one full trip or one period of time.

The equation for the frequency is (1/2pi) *(k/m)^0.5. So the inverse would be one whole period. This equation requires some reasoning to derive. Do they want you to memorize these things? And if you use this does it come out to the correct answer?

 

[pgardn]

The angular speed is related to f by the 1/2pi. A full trip is 2pi radians.

So they took omega, which is referred to the angular speed or frequency, which is equal to the square root of k/m and divided this by 4. This has to be derived though. I guess they expect you to memorize this. And I find it odd that they would just stick a Force in the problem and not give you more information about if that was an average force, the force exerted by the spring on the wall when the car came to a halt or something...

 

[MCATPhys]

that makes sense - the time it takes to reach maximum displacement should be 1/4 of the period...

the equation for the period (in the book) is:

(2pi * (m/k)^0.5)1/4

the answer comes to 0.287 seconds

the answer they show is:

((k/m)^0.5)1.4 = 1.37 seconds

this question is so confusing...

 

[pgardn]

that makes sense - the time it takes to reach maximum displacement should be 1/4 of the period...

the equation for the period (in the book) is:

(2pi * (m/k)^0.5)1/4

the answer comes to 0.287 seconds

the answer they show is:

((k/m)^0.5)1.4 = 1.37 seconds

this question is so confusing...

If possible could you type out the whole question. Getting bits and pieces makes it difficult to decipher. And we also need to know how frequency and angular speed or angular frequency are related. I need to show you this, but if you could give me the whole question it would be easier.

 

[pgardn]

that makes sense - the time it takes to reach maximum displacement should be 1/4 of the period...

the equation for the period (in the book) is:

(2pi * (m/k)^0.5)1/4

the answer comes to 0.287 seconds

the answer they show is:

((k/m)^0.5)1.4 = 1.37 seconds

this question is so confusing...

The angular frequency, that is how many radians one travels in 1 sec is related to how an object moves in a circle. If you look at just the linear part of an object moving in a circle at a constant angular frequency, omega, it mimics how the spring and the mass move in one dimension. So the equation they used was angular frequency = 2pi(1/T). And angular frequency is just equal to sqrt(k/m) this is what they were doing. Your problem would be 1/4 of this because you would theoretically just be making 1/4 of the total trip. The total trip would require that the car/spring would hit the wall, compress bounce back, go past the point where it original made contact, come to a halt, then come back till it reached the original postion just as it the spring makes contact with the wall.

 

[MCATPhys]

If possible could you type out the whole question. Getting bits and pieces makes it difficult to decipher. And we also need to know how frequency and angular speed or angular frequency are related. I need to show you this, but if you could give me the whole question it would be easier.

http://books.google.com/books?id=pPleL5NOtb4C&pg=PA313&lpg=PA313&dq=%22an+experiment+is+conducted+to+test+the+quality+of%22&source=bl&ots=wu9S0TTMAC&sig=3zfdh75hwXTktIpxre7r8QuO_-M&hl=en&ei=yMsCTLqMKZH0NfP3rTs&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q=%22an%20experiment%20is%20conducted%20to%20test%20the%20quality%20of%22&f=false

The question is on page 313, and the question I'm stuck on is number 3.

 

[MCATPhys]

it's a big a#$ question

 

[MCATPhys]

The angular frequency, that is how many radians one travels in 1 sec is related to how an object moves in a circle. If you look at just the linear part of an object moving in a circle at a constant angular frequency, omega, it mimics how the spring and the mass move in one dimension. So the equation they used was angular frequency = 2pi(1/T). And angular frequency is just equal to sqrt(k/m) this is what they were doing. Your problem would be 1/4 of this because you would theoretically just be making 1/4 of the total trip. The total trip would require that the car/spring would hit the wall, compress bounce back, go past the point where it original made contact, come to a halt, then come back till it reached the original postion just as it the spring makes contact with the wall.

that makes sense... but if angular frequency is the number of radians per second, then if we take one fourth of it.. that gives us the number of radians traveled in 1/4 of a second. i guess i don't understand become are final units will be Hz and not in seconds.

 

[pgardn]

that makes sense... but if angular frequency is the number of radians per second, then if we take one fourth of it.. that gives us the number of radians traveled in 1/4 of a second. i guess i don't understand become are final units will be Hz and not in seconds.

Ok this problem is completely diff from what I was envisioning based on your first few posts.

But to answer your question above. If an object is in SHM motion you can look at it using an object moving at a constant velocity moving in a circle of radius r. One full trip would around the circle would be 2pi radians or 2pi*r. If it went at a constant speed the object could be given a number that would represent how many radians it moves in each second. This is called omega. Now if you looked at this object traveling that was travening in a cirlce in 1 dimension, it would appear to just go back and forth (oscillate), you could give the maximum displacement to either side of the middle of the circle along a horizontal running right through the middle of the circle, like an x-axis. So it could be a given a frequency just like an object bobbing up and down on a spring. Which is just cylces per second. Well in the case of SHM using the mass moving along a circular path omega would equal 2pi*f.

Now which question were you answering on this car with springs on both sides? because there is a lot of conservation of energy and thermodynamics interwoven it this question. And I had no idea there were two springs and loops. Also, These circles or loops on the cars track have nothing to do with what I was discussing above.

 

[MCATPhys]

it's the third question...

 

[MCATPhys]

but why is it wrong to take 1/4 of the period?

 

[pgardn]

but why is it wrong to take 1/4 of the period?

Ok now I can see the problem. Question #3 states, What is the time difference between when the car passes the first circle and the time when it comes to a complete halt on the right wall...?

To begin with, if we assume the car is going 50 m/s at the 50m from the loop, and then goes through the loop and hits the wall and comes to a momentary halt, an then comes back to its orginal spot it has only gone through a half of its trip along the horizontal.

Does this make sense? One step at a time now so we don't get confused.

 

[MCATPhys]

It's a huge question, and I didn't want to confuse people...

but you're right... the car is maintaining a constant velocity of 50 m/s before and after the loop. And the distance from the end of the loop to the wall is 50 meters. So:

d = vt
t = d/v = 50/50 = 1 second.

To answer the question then, we have to add that one second to the time it takes for all the kinetic energy in the car to convert to elastic potential energy.

 

[pgardn]

It's a huge question, and I didn't want to confuse people...

but you're right... the car is maintaining a constant velocity of 50 m/s before and after the loop. And the distance from the end of the loop to the wall is 50 meters. So:

d = vt
t = d/v = 50/50 = 1 second.

To answer the question then, we have to add that one second to the time it takes for all the kinetic energy in the car to convert to elastic potential energy.

hold on just a second...

Did you get the number 9.13 meters from saying that the cars original kinetic energy is all converted to elastic potential energy? If so, this means the car/spring is going to meet the wall at a distance of 50 - 9.13 meters from the beginning of the loop, yes? or it will meet the wall before it goes the full 50 meters. So the first 40 something meters is at a constant velocity, but not the last 9.13 meters, yes?

 

[pgardn]

And I am sorry we had to go through all that other crapola. We did not need to for this question.

 

[MCATPhys]

I got that number my equalling 1/2mv^2 = 1/2kx^2 and solving for x...

I think the 50 meters the car travels doesn't include the above distance.

If you look at the answer in the back of the book, you'll understand what i mean.. one sec i'll get the link

 

[MCATPhys]

http://books.google.com/books?id=pPleL5NOtb4C&pg=PA313&lpg=PA313&dq=%22an+experiment+is+conducted+to+test+the+quality+of%22&source=bl&ots=wu9S0TTMAC&sig=3zfdh75hwXTktIpxre7r8QuO_-M&hl=en&ei=yMsCTLqMKZH0NfP3rTs&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q=then%20the%20speed%20of%20the%20car&f=false

look at answer 3 on the right column

page 366

 

[pgardn]

http://books.google.com/books?id=pPleL5NOtb4C&pg=PA313&lpg=PA313&dq=%22an+experiment+is+conducted+to+test+the+quality+of%22&source=bl&ots=wu9S0TTMAC&sig=3zfdh75hwXTktIpxre7r8QuO_-M&hl=en&ei=yMsCTLqMKZH0NfP3rTs&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q=then%20the%20speed%20of%20the%20car&f=false

look at answer 3 on the right column

page 366

Ok to me this is very poorly set up if you have shown me everything. So we will assume 50 meters is covered in 1s. And then the car/spring meets the wall and begins to slow down to a halt.

So now we must find the time it takes to accelerate till v final is 0 m/s. This occurs over a 9.13 meter distance. Using just the conservation of energy we will be unable to determine time. We must have some kinematics in this last part.

So here is how I see it unless there is something else not given. The force exerted on the car by the wall can be described as F = -kx... The negative is there because the force is in the opposite direction the car is moving. Is this ok so far, have you seen this before?

 

[pgardn]

Next F = -kx = ma

Now what this means is we can find the avg. acceleration (deceleration). If we have this, we can find the time it takes for a 100 kg car moving at 50 m/s to come to a halt by
Vf = Vo + at Vf = 0 m/s and a = -kx/m and Vo = 50 m/s. Solve for t... and I don't get their answer.

This is how I see it. I can't find the answer on pg 366?

 

[MCATPhys]

It's alright. This is such a stupid question. Thanks for spending so much time to help me.

 

[MCATPhys]

Here's the answer:

"if friction is negligible, then the speed of the car after coming out of the circle is the same as the speed when it went in, because no work was done on it. The distance to travel toward the wall is 50m, which can be traversed in 1 second. but the car will have to convert all of its kinetic energy into elastic potential energy before it stops. 1/4 (k/m)^0.5 = 1.3 s. Therefore, the total time is 2.3 s.