Calculating Spring Constant for Mechanical Oscillation

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Homework Help Overview

The problem involves calculating the spring constant for a spring that is used in mechanical oscillation. A 4.0 kg block extends the spring by 16 cm, and then a 0.50 kg body is hung from the same spring to analyze its oscillation period.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the mass when it is at rest and how these relate to the spring constant. There is mention of using conservation of energy and equilibrium conditions to find the spring constant.

Discussion Status

Some participants have confirmed the approach of using equilibrium conditions to relate the mass and spring constant. There is ongoing exploration of how to derive the period of oscillation from the spring constant.

Contextual Notes

Participants are working under the assumption that the spring behaves linearly and that the system is ideal. There is a focus on the relationship between mass, gravitational force, and spring extension.

-Aladdin-
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A 4.0kg block extends a spring 16cm from its unstretched position.The block is removed and a 0.50Kg body is hung from the same spring.If the spring is stretched and released,the period of oscillation is :

a)0.28s
b)0.02s
c)0.42s

My Work :

T = 2pi/w

T = 2pi*sqrt{Mass Totall/K}

I need to find K .

By applying law of conservaton of energy :

EMi = EMf

0.5K(16*10^-2)^2 = -mgh

h = 16*10^-2 ? ?

Any help please.
Thank you
 
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Try analyze which forces that are acting on the first mass when it hangs still extending the spring 16 cm. Think about what the sum of all these forces must be and how that will allow you to find k.
 
The very first sentence of the problem statement contains enough information to calculate the spring constant. From that you should be able to calculate the frequency and therfore, period using the 2nd mass.
 
So at equilibrium

mg = k*(16*10^-2)

Am I correct ?!
 
-Aladdin- said:
So at equilibrium

mg = k*(16*10^-2)

Am I correct ?!

Yes, that is correct.
 

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