Motion of an object attached to a spring

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A 5.0 kg mass attached to a spring stretches it by 10 cm, and when displaced an additional 5.0 cm and released, its motion can be modeled as X(t) = 0.05cos(10t). The amplitude of the motion is determined to be 0.05 m. Angular frequency (w) is calculated using the formula w = sqrt(k/m), where k is the spring constant derived from Hooke's Law (F = -kx). The force acting on the spring is equal to the weight of the mass, allowing for the calculation of k. This approach provides a complete understanding of the motion of the mass-spring system.
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Homework Statement


A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately?


Homework Equations


X(t) = Acos(wt+φ)


3. The Attempt at a Solution
the awnser to this problem is 0.05cos10t. I see where they get the 0.05 from since it is the amplitude but how do they figure out w(angular frequency)=10
 
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w = sqrt (k/m)

F = -kx

Now you have the equations you need
 
how do i figure out F acceleration is not given
 
F = -kx

is Hooke's Law.

The force comes from the weight of the body. And x is the extension of a spring.


Use this to find k.

Then apply the other formula to find w.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

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