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

[Tahir Mushtaq]

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.

 

[kyiydnlm]

$$\frac{L^2}{4} \theta K+J\ddot{\theta} = 0$$
$$J = \frac{1}{12}ML^2$$
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.

 

[tomcue]

In both cases, we can use Newton's second law of motion to derive the differential equation for the spring-mass system.

Case 1:

Let x(t) be the displacement of the mass from its equilibrium position, and let y(t) be the angular displacement of the seesaw from its balanced position. The pivotal point will be the origin of our coordinate system.

The forces acting on the mass are the spring force (k1x) and the force of gravity (mg). The force acting on the seesaw is the torque due to the mass (mx) and the torque due to the spring (k1xy).

Using Newton's second law for the mass:

mx'' = -k1x - mg

And for the seesaw:

Iy'' = -k1xy

Where I is the moment of inertia of the seesaw.

We can also use the relationship between the angular and linear displacements (y = x/l), where l is the length of the seesaw:

Iy'' = mlx''

Substituting this into the equation for the seesaw, we get:

mlx'' = -k1xy

Combining this with the equation for the mass, we get the differential equation for the spring-mass system attached to one end of a seesaw:

mx'' + k1x + mg = 0

Case 2:

In this case, we have two springs (with spring constants k1 and k2) attached to both ends of the seesaw. The forces acting on the mass are now the sum of the forces from both springs (k1x + k2x) and the force of gravity (mg). The forces acting on the seesaw are the sum of the torques from both springs (k1xy + k2xy) and the torque due to the mass (mx).

Using Newton's second law for the mass:

mx'' = -(k1 + k2)x - mg

And for the seesaw:

Iy'' = -(k1 + k2)xy

Substituting the relationship between angular and linear displacements, we get:

mlx'' = -(k1 + k2)x

Combining this with the equation for the mass, we get the differential equation for the spring-mass system attached to both ends of a seesaw:

mx'' + (k1 + k2)x + mg = 0

In both cases, we can solve these second-order differential equations