Calculating Time for a Mass Hitting a Spring

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The discussion centers on calculating the time it takes for a mass to come to a stop after colliding with a spring on a frictionless surface. The initial approach involved using Hooke's law to find force and acceleration, but it was clarified that the relationship is not linear, and the force versus time graph is sinusoidal due to simple harmonic motion (SHO). The correct method involves solving the equation of motion, leading to the conclusion that the time to stop is one-fourth of the period of the oscillation. The angular frequency is determined as ω = √(k/m), and the time to stop can be expressed as t = π/(2ω). This understanding is essential for effectively teaching the concept in a classroom setting.
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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 (vf = 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
 
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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 /> <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.
 
ninevolt said:
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 /> <br />

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

AlephZero said:
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
 
ninevolt said:
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 /> <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.
 
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eggman1965 said:
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!
 
AlephZero said:
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
 
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