MHB Simple Harmonic Motion: 5kg Particle Suspended by 500 N/m String

[markosheehan]

A particle of mass 5 kg is suspended from a fixed point by a light elastic string
which hangs vertically. The elastic constant of the string is 500 N/m.
The mass is pulled down a vertical distance of 20 cm from the equilibrium
position and is then released from rest.
(i) Show that the particle moves with simple harmonic motion.

 

[I like Serena]

A particle of mass 5 kg is suspended from a fixed point by a light elastic string
which hangs vertically. The elastic constant of the string is 500 N/m.
The mass is pulled down a vertical distance of 20 cm from the equilibrium
position and is then released from rest.
(i) Show that the particle moves with simple harmonic motion.

Hi markosheehan! ;)

An elastic string provides a force that is linear with the displacement from the rest position.
As a consequence we have simple harmonic motion.
If we want to prove it, we have to set up the equation of motion and solve it.
Care to give it a try?

For the record, we expect some kind of attempt or explanation where you're stuck, if only to figure out how to help you best.

 

[markosheehan]

what i tryed to do was find the force down and the force up find the resultant force and let it equal to F=5a and then that would prove it but to do this when i am finding the force in the string i need the natural length of the string but it is not given in the question. to find the force up i use F=k(length-natural length)

 

[I like Serena]

what i tryed to do was find the force down and the force up find the resultant force and let it equal to F=5a and then that would prove it but to do this when i am finding the force in the string i need the natural length of the string but it is not given in the question. to find the force up i use F=k(length-natural length)

We don't need the natural length.
When the string has its natural length, the elastic force is zero.
The elastic force is proportional with the change in its natural length.

So let's start with the position in equilibrium.
Then the force of gravity is equal and opposite the elastic force.
That is:
$$F_g = F_e \quad\Rightarrow\quad mg = k\Delta y \quad\Rightarrow\quad 50 = 500 \Delta y\quad\Rightarrow\quad \Delta y = 0.10 \text{ m}$$

Now initially we pull the mass $20 \text{ cm}$ down and we let go.
Then we have:
$$F_{net} = F_e - F_g = ky-mg = 500y-50 = 500(y-0.10)$$

In other words, the net force is proportional to the distance from the equilibrium position.