Understanding spring mass system

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The discussion focuses on the dynamics of a spring-mass system, where a mass is displaced by a force F, causing the spring to stretch a distance X. The equations derived include the static force balance f - kx = net force and the dynamic equation f - kx - ma = 0, which applies Newton's second law. It is confirmed that the equations are correct as long as the force is continuously applied and the mass is not released. If the mass is in equilibrium, then f - kx = 0; otherwise, it will accelerate. Understanding these principles is essential for mastering basic dynamics in spring-mass systems.
chandran
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I am just learining basic dynamics from spring mass system.
there is a horizontal spring fixed to the wall and a mass is at the end.
When the mass is displaced by a force F to the right the spring stretches to a distance X. Now i draw the STATIC FBD. The
eqn is f-kx=net force on the mass. Now i draw the kinetic diagram and the equation i derive is

f-kx=ma where a is the acceleration of the mass m. So for this spring mass system the dynamic equation is
f-kx-ma=0.

Is what i have done is true and correct?
 
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Is the force acting continuously on the object...?Or it just stretches the string & let's it go to oscillate...?

That equation,vector or scalar,f-kx-ma=0 makes no sense...

Daniel.

Daniel.
 
chandran said:
I am just learining basic dynamics from spring mass system.
there is a horizontal spring fixed to the wall and a mass is at the end.
When the mass is displaced by a force F to the right the spring stretches to a distance X. Now i draw the STATIC FBD. The
eqn is f-kx=net force on the mass. Now i draw the kinetic diagram and the equation i derive is

f-kx=ma where a is the acceleration of the mass m. So for this spring mass system the dynamic equation is
f-kx-ma=0.

Is what i have done is true and correct?

Yes, it is correct as long the hand pulling the mass is not releasing it.

F - kx = 0
 
chandran said:
I am just learining basic dynamics from spring mass system.
there is a horizontal spring fixed to the wall and a mass is at the end.
When the mass is displaced by a force F to the right the spring stretches to a distance X. Now i draw the STATIC FBD. The
eqn is f-kx=net force on the mass.
This is just a statement that the net force on the mass equals f - kx.
Now i draw the kinetic diagram and the equation i derive is

f-kx=ma where a is the acceleration of the mass m. So for this spring mass system the dynamic equation is
f-kx-ma=0.
Now you've applied Newton's 2nd law: Net Force = ma. Note that rewriting it as "Net Force - ma = 0" adds nothing.

If the mass is in equilibrium, then f - kx = 0; if not, then it will accelerate.
 
chandran said:
I am just learining basic dynamics from spring mass system.
there is a horizontal spring fixed to the wall and a mass is at the end.
When the mass is displaced by a force F to the right the spring stretches to a distance X. Now i draw the STATIC FBD. The
eqn is f-kx=net force on the mass. Now i draw the kinetic diagram and the equation i derive is

f-kx=ma where a is the acceleration of the mass m. So for this spring mass system the dynamic equation is
f-kx-ma=0.

Is what i have done is true and correct?

Here, you can learn more on basic dynamics. make sure that you are able to solve the given exercises...

https://www.physicsforums.com/showthread.php?t=72040


marlon
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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