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Sprin block system with ideal spring constant k & block of mass m.The block can oscillate simple harmonically in vertical direction & also spring-block can oscillate in a small circular arc like a pendulum & due to both the oscillations block moves on a line forming letter 8.Find the lenght of the spring during oscillation.

Hello Aditya,

I can't answer that with only k and m. In order for the path of the mass to describe an 8, the mass must have been displaced from its equilibrium position, point E, to a point D. And then as it was released, it must have been given a velocity vector, V, perpendicular to a line through points D and E. The system's equilibrium point is at the point where the 8-shaped path of the mass crosses itself. Let's assign an xy coordinate system to the system's motion such that the line through D and E is along the y axis. Let point E be at the xy system's origin. In that case the velocity V must have been in the x axis. Let the distance between D and E be called de.

The kinetic energy of the mass as it passes through the equilibrium point is equal to the energy given to the system both in the displacement to point D and the velocity perpendicular to DE. But the x axis energy and the y axis energy are independent. The y axis energy was given to the system in stretching (or compressing the spring) to point D. The x axis energy was given to the system in giving the mass a horizontal velocity as it was released.

So the x axis motion is not a factor in the changes in the spring's length. The change in the spring's length, as measured in the +y and -y direction, is 2*de.

I hope this helps,

Steve

Physics

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