Question

One end of a spring with a force constant of k = 10.0 N/m is attached to the end of a long horizontal frictionless track and the other end is attached to a mass m = 2.20 kg which glides along the track. After you establish the equilibrium position of the mass-spring system, you move the mass in the negative direction (to the left), compressing the spring 2.48 m. You then release the mass from rest and start your stopwatch, that is x(t = 0) = −A, and the mass executes simple harmonic motion about the equilibrium position. Determine the following.(a) displacement of the mass (magnitude and direction) 1.0s after it is released
(b) velocity of the mass (magnitude and direction) 1.0s after it is released
(c) acceleration of the mass (magnitude and direction) 1.0s after it is released
(d) force the spring exerts on the mass (magnitude and direction) 1.0s after it is released
(e) How many times does the object oscillate in 12.0s?

Answer 1

Answer:

-2.478

0.379

11.14

24.78

Explanation:

Angular frequency of spring in harmonic motion is given by?

ω = √(k/m)

ω = √(10/2.2)

ω = √4.54

ω = 2.13 s^-1

If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine

x(t) = -A * cos(ωt)

x(t) = -2.48 * cos(2.13 * 1)

x(t) = -2.48 * 0.9993

x(t) = -2.478

Velocity and acceleration are 1st and 2nd derivative of position

b)

v(t) = Aω * sin(ωt)

v(t) = 2.48 * 2.13 * sin(2.13 * 1)

v(t) = 5.282 * sin2.13

v(t) = 5.282 * 0.03717

v(t) = 0.379 m/s

c)

a(t) = Aω^2 * cos(ωt)

a(t) = 2.48 * 2.12² * cos(2.13 * 1)

a(t) = 2.48 * 4.494 * cos2.13

a(t) = 11.15 * 0.9993

a(t) = 11.14 m/s²

d)

F = -k * x(t)

F = -10 * -2.478

F = 24.78 N