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1

My explanation is as follows Force on the spring in stretching it to length x can be written as $F=k(x-L_0)$ where x is the displacement, $L_0$ is the initial length of the spring and k is spring constant. energy stored is $dE=F.dx$ upon integration we will get $E=\frac{1}{2}.k.(L-L_0)^2$ where $L$ is the final length of the spring. This is I think ...


2

Precisely, the point is that the slope is not equal to 1. The ratio $\dfrac{F}{\Delta L}=k$. Therefore, $Z=\dfrac{1}{2}(x)*(\dfrac{x}{k})$ $= \dfrac{1}{2k}x^2$. Which looks wrong but is true because in the notation you are using, $x$ is not the elongation but is rather the force acting. So in a more familiar notation $E=\dfrac{1}{2k}F^2$. If you want to ...


0

Is classical mechanics valid when a force is applied on one end of a massless spring(assuming we can have a massless spring, or can't we? "yes we can") while the other end is grounded" yes is valid "will there be an oscillation about new position(F/k) if the equations of motion are still valid." correct Can the physical behavior be plotted? yes , sure


2

Start with a conventional mass on a massless spring and see what happens if we let the mass go to zero. The equation of motion for the mass $m$ on a spring with force constant $k$ is: $$ x = A\sin\left(\sqrt{\frac{k}{m}}t\right) + B\cos\left(\sqrt{\frac{k}{m}}t\right) $$ where we get the constants $A$ and $B$ from the velocity and position at time $t=0$. ...


-1

Yes, spring mass will increase but in very very small amount. As per Mass energy equivalence the amount of mass gain will be(E=mc2): m= E/C2 for example if spring have store 1 Joule of potential energy when it is extended then the increased mass will be: m =1/299792458= 3.33564095198152e-9



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