Differential equation of spring-mass system attached to one end of seesaw

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SUMMARY

The discussion focuses on deriving the differential equations for a spring-mass system attached to a seesaw. In Case 1, where a spring with spring constant K1 is attached to a balanced seesaw with total mass M, the oscillation occurs around the pivotal point. The differential equation for this system is not explicitly stated but is implied to be related to the spring constant and mass. In Case 2, where springs with constants K1 and K2 are attached to both ends of the seesaw, the equation is given as \(\frac{L^2}{4} \theta K + J\ddot{\theta} = 0\), where \(J = \frac{1}{12}ML^2\) and \(K = K1 + K2\). The discussion emphasizes the importance of the seesaw's uniformity and pivot location on the resulting equations.

PREREQUISITES
  • Understanding of differential equations
  • Knowledge of spring constants (K1, K2)
  • Familiarity with rotational dynamics and moment of inertia (J)
  • Basic principles of oscillatory motion
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  • Study the derivation of differential equations for spring-mass systems
  • Learn about the effects of varying spring constants on oscillation
  • Research the principles of rotational dynamics in mechanical systems
  • Explore the impact of pivot points on the motion of seesaws
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Tahir Mushtaq
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please help me to find differential equation of spring-mass system attached to one end of seesaw.

case 1: Seesaw is balanced with its fulcrum point or pivotal point. At one end of seesaw, the spring (with spring constant K1) is attached. Now the seesaw has total mass M which is attached to spring, form a spring-mass system. I confuse on this point that the oscillation of seesaw will be around pivotal point. Then what is differential equation of spring-mass system.

case 2: The same case with both ends of Seesaw is attached with springs having spring constant K1 and K2. Then what is differential equation of this system.
 
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[tex]\frac{L^2}{4} \theta K+J\ddot{\theta} = 0[/tex]
[tex]J = \frac{1}{12}ML^2[/tex]
case 1: K = K1
case 2: K = K1+K2

Assuming uniform seesaw, pivot is center of mass (also geometry center). If otherwise seesaw is not uniform, J will change accordingly. If pivot is not geometry center, the equation will be slightly different.
 

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