Derivation of the oscillation period for a vertical mass-spring system

[User1265]

Homework Statement

I am confused how the time period of a vertical spring system is T= 2π√m/k, when there is the weight of the mass to consider?

Relevant Equations

T= = 2π *√m/k

I understand the derivation of T= 2π√m/k is a= -kx/m, in a mass spring system horizonatally on a smooth plane,
as this equated to the general equation of acceleration of simple harmonic motion , a= - 4π^2 (1/T^2) x

but surely when in a vertical system , taking downwards as -ve, ma = kx - mg , so I don't understand why in the following context of the question below

A bungee jumper of mass 70kg has a light linear elastic bungee cord with coefficient of elasticity 153N/m. He jumps from the platform and then bounces up and down at the end of the bungee cord. Assuming that the amplitude of bouncing is small enough that the bungee cord is never slack, what is the period of the steady state bouncing?

Why is T = 2π *√m/k is used to calculate the period, surely it can't have the same derivation as a horizontal spring-mass system as there is the constant force of gravity to consider?

My solution was intially to use:
Initially I used T = 2π *√m/k, but I had never really considered if I could get to the same derivation using a vertical spring system until now.

 

[haruspex]

surely it can't have the same derivation as a horizontal spring-mass system as there is the constant force of gravity to consider?

Gravity adds a constant downward force. The initial extension of the rope, i.e. to the equilibrium position, adds an equal and opposite force. We can take that as constant and just consider variations from there.
During oscillation, the change in the tension is proportional to the change in extension.

 

[jtbell]

Suppose m = 1 kg and k = 100 N/m. If you need to calculate gravitational force, let the magnitude of g = 10 m/s² for simplicity in calculations.

Place the spring+mass system horizontally on a frictionless surface. When the mass is at its equilibrium position (x = 0), F = 0. When the mass is at x = -0.01 m (to the left of the equilbrium position), F = +1 N (to the right). When the mass is at x = +0.01 m (to the right of the equilibrium position), F = -1 N (to the left).

Now place the spring+mass system vertically, with the spring's anchor point at the top. The spring stretches and settles into a new equilibrium length. Call the mass's equlibrium position x = 0, with x positive upwards and negative downwards.

What is the net force on the mass at x = 0?

What is the net force on the mass at x = -0.01 m (below the equilibrium position)?

What is the net force on the mass at x = +0.01 m (above the equilibrium position)?

 

[User1265]

Gravity adds a constant downward force. The initial extension of the rope, i.e. to the equilibrium position, adds an equal and opposite force. We can take that as constant and just consider variations from there.
During oscillation, the change in the tension is proportional to the change in extension.

So what would be considered the restoring force? Is it just the tension or the resultant force which would be acting towards the equilbrium?

 

[BvU]

As the term 'restoring force' indicates: it acts towards equilibrium, so it must be the resultant force (the F_net in F = ma ).

 

[User1265]

As the term 'restoring force' indicates: it acts towards equilibrium, so it must be the resultant force (the F_net in F = ma ).

But doesn't the elastic force in the cord acts towards equilibrium too ?

 

[BvU]

No. It acts towards the natural length of the cord.

 

[User1265]

No. It acts towards the natural length of the cord.

Thank you, and the restoring force proportional to -ve displacement?

 

[User1265]

Gravity adds a constant downward force. The initial extension of the rope, i.e. to the equilibrium position, adds an equal and opposite force. We can take that as constant and just consider variations from there.
During oscillation, the change in the tension is proportional to the change in extension.

Why change in tension and not tension is proportional to displacement at an instant in the elastic cord as the cord is originally displaced from its natural length?

 

[haruspex]

Why change in tension and not tension is proportional to displacement at an instant in the elastic cord as the cord is originally displaced from its natural length?

Tension is proportional to the extension, of course, but that does not lead to an understanding of why gravity makes no difference to the oscillation.
The relevant point is that in the equilibrium position the tension balances gravity, and extension relative to that position is proportional to the change in tension. The change in tension becomes the net force.