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

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The discussion focuses on deriving the differential equations for a spring-mass system attached to a seesaw in two scenarios. In the first case, a single spring with spring constant K1 is attached to one end of a balanced seesaw with mass M, leading to oscillations around the pivot point. The second case involves two springs with constants K1 and K2 attached to both ends of the seesaw, resulting in a combined spring constant K equal to K1 + K2. The equations of motion are influenced by the seesaw's mass distribution and pivot position, with the moment of inertia J calculated as J = (1/12)ML^2 for a uniform seesaw. The discussion emphasizes that variations in the seesaw's uniformity and pivot location will alter the resulting differential equations.
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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\frac{L^2}{4} \theta K+J\ddot{\theta} = 0
J = \frac{1}{12}ML^2
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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