Calculating Time for a Mass Hitting a Spring

[eggman1965]

I posed a question to my gr12 physics class today along these lines:

suppose a mass (m) traveling on a frictionless surface at a speed (v) runs into a spring (constant K). How long (time t) will it take to come to a stop (v_f = 0)

I'm quite rusty I suppose since I cannot seem to come up with the function that describes the F vs t graph. Finding the distance is a piece of cake (1/2kx²=1/2mv²) but I'm stuck trying to find the time. Can anyone help?

Thanks

 

[ninevolt]

You have the initial velocity, and you have the force applied to the mass.
Therefore you can find the acceleration of the mass.

F = ma = -kx (by hooke's law)

The sum of the total force applied is equal to:
$$F = \int_{0}^{x}-kx\hspace{5} dx<br />$$

then just divide F by m
find acceleration
then you have the time it takes for Vinitial to change into Vfinal (0)

 

[AlephZero]

Here's a reminder: Simple harmonic motion.

 

[eggman1965]

You have the initial velocity, and you have the force applied to the mass.
Therefore you can find the acceleration of the mass.

F = ma = -kx (by hooke's law)

The sum of the total force applied is equal to:
$$F = \int_{0}^{x}-kx\hspace{5} dx<br />$$

then just divide F by m
find acceleration
then you have the time it takes for Vinitial to change into Vfinal (0)

Here's a reminder: Simple harmonic motion.

This does not give the force vs time relation which is NOT linear. The F vs t graph is either quadratic or exponential and the standard kinematics equations only apply when acceleration is constant (which it is NOT in this case) The S.H.O. is my current focus but I'm still trying to figure out a way to present this to my class in such a way that they can see it.

Thanks Anyway

 

[nasu]

You have the initial velocity, and you have the force applied to the mass.
Therefore you can find the acceleration of the mass.

F = ma = -kx (by hooke's law)

The sum of the total force applied is equal to:
$$F = \int_{0}^{x}-kx\hspace{5} dx<br />$$

then just divide F by m
find acceleration
then you have the time it takes for Vinitial to change into Vfinal (0)

The equation is not correct. The integral is not a force but the work of the elastic force.
Dividing it by m will not give acceleration.

In order to find the position, velocity, etc versus time you need to solve the equation of motion.
In this case it is
F/m=a
or -kx/m=d2x/dt2
(F=-kx, a=d2x/dt2)
This is the equation of a SHO and we already know the general solution. So we know that the motion will be a SH motion and can apply the results.

Once the mass connects with the spring, you'll have a SHO.
The initial position corresponds to the zero displacement. It will keep moving until v=0. At this point it will start moving backwards. So the time you are looking for is the time to move between the equilibrium and an extreme, or 1/4 of a period.

Note: edited to correct error. It is 1/4 period and not 1/2 as I wrote originally.

 

[AlephZero]

This does not give the force vs time relation which is NOT linear. The F vs t graph is either quadratic or exponential

No, it is sinusoidal.

The angular frequency $$\omega = \sqrt{k/m}$$

Measuring time from when the mass hits the spring,

$x = A \sin \omega t$ for some constant A.

$\dot x = A\omega \cos \omega t$

When t = 0 you have $\dot x = A \omega = v$ so $A = v/\omega$

The time to stop is independent of the velocity and is 1/4 of a complete cycle, so

$\omega t = \pi/2$ or $t = \pi / (2 \omega)$

and the standard kinematics equations only apply when acceleration is constant (which it is NOT in this case) The S.H.O. is my current focus but I'm still trying to figure out a way to present this to my class in such a way that they can see it.

Hmm... that could be a tough challenge!

 

[eggman1965]

No, it is sinusoidal.

The angular frequency $$\omega = \sqrt{k/m}$$

Measuring time from when the mass hits the spring,

$x = A \sin \omega t$ for some constant A.

$\dot x = A\omega \cos \omega t$

When t = 0 you have $\dot x = A \omega = v$ so $A = v/\omega$

The time to stop is independent of the velocity and is 1/4 of a complete cycle, so

$\omega t = \pi/2$ or $t = \pi / (2 \omega)$



Hmm... that could be a tough challenge!

Thank you! This is what I was looking for, and yes my rust is showing as it is of course a sine function not a quadratic/exponential as I erroneously stated earlier.

Sincerely

eggman1965