Asked by Eden
A 0.70 kg mass at the end of a spring vibrates 4.0 times per second with an amplitude of 0.15 m.
Determine the velocity when it passes the equilibrium point.
Determine the velocity when it is 0.11 from equilibrium.
Determine the total energy of the system.
Determine the velocity when it passes the equilibrium point.
Determine the velocity when it is 0.11 from equilibrium.
Determine the total energy of the system.
Answers
Answered by
Eden
sorry, .11 m from equilibrium.
Answered by
bobpursley
start with the basic equations.
x=A*sinwt
v= Aw*coswt
a= -Aw^2sinwt
at the equilibrium point wt is PI (cosine is max), so V-Aw you know A, and w=2PI/Period, and period is given (.25sec).
when position is .11,
.11=Asinwt solve for wt, then v= Awcoswt
energy? 1/2 m v^2 where v is vmax, or Aw
x=A*sinwt
v= Aw*coswt
a= -Aw^2sinwt
at the equilibrium point wt is PI (cosine is max), so V-Aw you know A, and w=2PI/Period, and period is given (.25sec).
when position is .11,
.11=Asinwt solve for wt, then v= Awcoswt
energy? 1/2 m v^2 where v is vmax, or Aw
Answered by
Eden
what is Pl?
Answered by
bobpursley
PI is 3.14159...
Answered by
Eden
oh... pi... thanks
Answered by
Anonymous
how do you know period is .25 sec?
Answered by
Anonymous
frequency is four times per second! (1 sec divided by four times is .25)
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