Time period of a spring system

[joeyjo100]

The equation for the time period of a mass spring system does not contain gravitational field strength as a variable, implying that the frequency of oscillations is independent of gravity, and that the time period of a mass sping system would be the same on Earth a on the moon, for example.

But surely gravity could affect the system, as it will exert a force on the spring system, maing it accelerate more.

Does a mass-spring system have the same time period whether it is perpendicular to a gravitational field or horizontal?

 

[nasu]

Gravity influences the equilibrium position of the system but not its period, at least for ideal spring.
So the period is the same for horizontal (without friction) and vertical system.

 

[AlephZero]

But surely gravity could affect the system, as it will exert a force on the spring system, maing it accelerate more.

The weight of the mass (i.e. the gravitational force) is constant. The extra force in the spring, measured about its new equilibrium position, is equal and opposite to the weight so they cancel out.

Does a mass-spring system have the same time period whether it is perpendicular to a gravitational field or horizontal?

They are the same. The general solution of

m\ddot x + kx = F[/itex]<br /> <br /> where F is a constant force (for example weight) is<br /> <br /> x = A \cos \omega t + B \sin \omega t + F/k [/itex]&lt;br /&gt; &lt;br /&gt; where \omega^2 = k/m.&lt;br /&gt; &lt;br /&gt; The constant term F/k does not change the period of the oscillation.