A mass m slides without friction on a 1-dimensional track; the initial velocity

Question

A mass m slides without friction on a 1-dimensional track; the initial velocity is v0 as shown. At time t = 0 the mass contacts the end of the spring (at position x = 0). The force exerted by the spring is F(x) = ? k x.

(A) Determine the position where the mass will come to rest.

(B) Determine the time when the mass will come to rest.

Express the answers as formulas, in terms of the parameters in the figure.

Explanation / Answer

I can't see the image. I'll just say the mass is moving to the right, coming from negative x-values to x=0 and coming to rest at a distance of x. a(t)=(-k/m)*x(t), then from F=ma, and showing the functional dependencies of a and x. x''=(-k/m)*x Which has general solution: x(t)=A*cos(wt)+B*sin(wt) where w=sqrt(k/m) Differentiate: v(t)=-wA*sin(wt)+wB*cos(wt) x(0)=0 v(0)=+v0 Apply initial conditions 0=A*cos(0)+B*sin(0) ===> A=0 v0=-wA*sin(0)+wB*cos(0)=0+wB ===> B=v0/w x(t)=(v0/w)*sin(wt) v(t)=v0*cos(wt) The time at which it stops is when v(t)=0. 0=v0*cos(wt), wt=Pi/2, 3Pi/2, 5Pi/2..., infinitely many values since it oscillates. The first time it stops though is t=(Pi/2w)=(Pi/2)*sqrt(m/k) x(Pi/2w)=v0/w*sin(Pi/2)=v0/w=v0*sqrt(m/k) You can check this with energy considerations: 0.5*mv0^2=0.5*k*x^2 x=v0*sqrt(m/k), but energy won't get you the time dependence.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.