You are watching an object that is moving in SHM. When the object is displaced 0

Question

You are watching an object that is moving in SHM. When the object is displaced 0.630m to the right of its equilibrium position, it has a velocity of 2.10 m/s to the right and an acceleration of 8.20 m/s^2 to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Explanation / Answer

ma=kx a=(k/m)*x ==> a=8.2=(k/m)*0.63 k/m=13.01 s^-2 0.5*m*v^2+0.5*k*x^2=0.5*k*A^2 ...the kinetic and potential energy at this intermediate point is equal to the total potential energy at the farthest amplitude. You are solving for (A-x) Divide by 0.5 and divide by k (m/k)*v^2+x^2=A^2 A=sqrt(x^2+(m/k)v^2) A=sqrt(0.63^2+(1/13.01)*(2.1)^2) A=0.8578 m A-x=0.8578-0.63= 0.228m

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