Recent content by Thomas Brown

Yup, this makes sense! Thanks for your help!

I think I got it but can you please verify? I know that a = w^2 x is the acceleration in SHM, so I integrated the A(t) formula to get V(t) = xw * sin(wt) and velocity is greatest when sin(wt) = 1 so the maximum velocity is xw, or v = x * (k / m)^1/2.

I see that delta V / delta T would be the acceleration and that the velocity would be greatest when acceleration is zero (maximum) but if I divide both sides by delta T I end up with: -kx = m dV/dT -kx = ma Fn = ma And I'm not really sure what I could do next...

1. A mass M is connected to a wall by an ideal spring. The mass is on a frictionless surface. The mass is pushed toward the wall, compressing the spring by a distance X. Use the impulse-momentum theorem to demonstrate that the mass will reach a maximum velocity of X * (K / M)^1/2. 2. J = delta...