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.
7 answers
sorry, .11 m from equilibrium.
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
what is Pl?
PI is 3.14159...
oh... pi... thanks
how do you know period is .25 sec?
frequency is four times per second! (1 sec divided by four times is .25)
0 answers