Question

a mass of 64.5g sits at the end of a spring on a horizontal frictionless table and undergoes simpel harmonic motion. The amplitude of this motion is 8.00m and the particle is at +8.00 when t=0s. the force at this time is -15.7N

1) What is the velocity at t=0.648

2) Whats the positon at this time

3) whats the acceleration at this time

4) what is the force constant of the spring

5) what is the first time other than 0.648 that the object has the maximum magnitude of the velocity

*include sign (-/+) when needed

Answer 1

mass = 64.5 g

amplitude = A = 8 m

At t= 0s particle is at 8 m i.e. the particle start SHM from the extended end.

force at t = 0 is -15.7 N

Force in a spring = -Kx = -K*8 = -15.7

4). K = 1.96 N/m.

The total energy will be constant = (1/2)KA² = (1/2)Kx² + (1/2)MV²

V =

X at t = 0.648 s = Acos(wt)

w =

2). X = A cos(1.14*0.648) = 8* cos(0.739) = + 5.92 m.

1). Velocity =

3). Accleration = Force/mass = Kx/M = (1.96*5.92)/0.0645 = 179.9 m/s²

Maximum velocity will occur at x = 0

X = Acos(wt) = 0

wt = pi/2

5). t = pi/(2w) = 3.14/(2*1.14) = 1.38 s.

Thank You !!!