SatmeerB

Quiz 3 1) m=1000 kg k= 5*10^6 x=3.16 v=0 In this problem the conservation of energy is the best to use, the car's kinetic energy is converted to potential when it hits the wall. Ebefore=Eafter 1/2*Kx^2 + 1/2*mvi^2=1/2Kx^2 + 1/2mvf^2, since there is no potential energy before and no kinetic energy after the equation becomes this:

1/2*mvi^2=1/2Kx^2 Plug in for the known values and sovle for vi

2) i.First the value of amplitude must be converted to meters instead of cm. And the velocity from cm/s^2 to m/s^2. The mass of the block is best found by using the maximum velocity equation (Vm) which equals omega* amplitude but since the velocity is being all the data is for when the block is halfway between the equilibrium and the endpoint the velocity measured must be multiplied by 2. Because the max velocity is always at the equilibrium point. So, Vm=Vgiven*2 now that we have Vm, we can plug in for Vm and amplitude in the Vm= omega*amplitude eqaution and solve for omega. After we have omega we can use the omega=sqrt(k/m) equation and solve for m. Ending up with m=k/omega^2. ii. The period (T) is given by the equation T=2pi*sqrt(m/k) T is found by simply plugging in for m which we solved for in the first part and plugging in for k which is given. iii. The maximum acceleration equation is Am= omega^2 *Xm omega= 2pi*T, T was just solved for and Xm is given in the original question. 3) i. To get the maximum speed of the block you must first get omega. Omega=sqrt(k/m). Both values k and m are given so omega can be solved for. Using omega you use the maximum velocity equation which is Vm= omega*amplitude. Omega was just solved for and amplitude was given in the question,

ii. The value of the distance from equilibrium must be converted to meters instead of cm. Getting the speed of the block from a distance from the equilibrium is done first by getting time(t). Using the displacement equation of x(t)=Xm*cos(omega*t+phi). Since phi=0 it goes away and now you plug in the distance from equilibrium for x(t) and solve for t. Once you have t you plug in the value in the velocity equation v(t)= -omega *Xm*sin(omega*t). Using omega from part one, t was just solved for and amplitude from the question.

iii. The maximum acceleration is is solved for by using the same value for t from the 2nd part of the question only this time plugging into the acceleration equation a(t)=-omega^2*Vm*cos(wt).