a = .09 meter
w = 2 pi f = sqrt (k/m) = sqrt(2.1/.9) = 1.53 radians/s
x = a sin w t
x = .09 sin 1.53 t
v = .09*1.53 cos 1.53 t = .1377 cos .153 t
That is max when cos = 1 so part a is .1377 m/s
a = -.09*1.53^2 sin wt = - 1.53^2 x
when x = .045
sin 1.53 t = .045/.09 = .5
so 1.53 t = pi/6 radians (30 degrees)
then t = .342 seconds (answer to part d by the way)
cos 30 degrees = .866
so
v = .1377*.866 = .119 m/s (part b)
a = -1.53^2*.09 * sin 30 = .105 m/s^2
Question 4
A 0.9 kg block attached to a spring of force constant 2.1 N/m oscillates with an amplitude of 9 cm. A)Find the maximum speed of the block. Answer in units of m/s
B)Find the speed of the block when it is 4.5 cm from the equilibrium position.
Answer in units of m/s
C)Find its acceleration at 4.5 cm from the equilibrium position. Answer in units of m/s^2
D)Find the time it takes the block to move from x = 0 to x = 4.5 cm. Answer in units of s
3 answers
Wow thank you. I was really lost.
this is a huge help! thanks again!