Hooke's law says that the return force for a spring operating in its linear region is proportional to the displacement from its equilibiurm length (F=-kx). If a block is attached to such a spring, pulled/pushed from equilibrium (x=0) and allowed to slide back and forth (frictionlessly), it will undergo simple harmonic motion. That is, its displacement will be of the form x(t)= x(sub0)cos(2pi t/T) where A is the initial displacement, t is the time, and T is the period. Using the relationship between position, velocity, and acceleration together with Newton's Second law (F=ma)

Question

anonymous · Answer

Find the relationship between the spring constant k and the mass m the initial displacement (xsub0) and the period T

jim_thompson5910 · Answer

I'm not that great at physics, so I could be way off, but here's what I think is the way to go


F = -kx .. Hooke's Law

ma = -kx ... plug in F = ma (Newton's second law)

ma = -k*x(t) ... rewrite x as a function of t

ma = -k*x_0*cos(2pi*t/T) ... plug in the position function given above

a = [-k*x_0*cos(2pi*t/T)]/m ... divide both sides by m


I think this is valid, but I don't know if this is what they want or not.

ganeshie8 · Answer

is this from calculus ?

ganeshie8 · Answer

if so, then a = x''
and u need to eliminate a

jim_thompson5910 · Answer

oh good point

anonymous · Answer

yes this is for calc

ganeshie8 · Answer

F = -kx

F  = ma

ma  = -kx

m $\frac{d^2x}{dx^2}  = - kx = 0$ ----------------(1)

ganeshie8 · Answer

you're given 
$x = x_0\cos (2\pi t/T)$ ----------(2)

can u find the second derivative of this ?

ganeshie8 · Answer

$x_0$ is just a constant above

ganeshie8 · Answer

find dx/dt and d^2x/dt^2, can u ? :)