Question

At t = 0, a 755 g mass at rest on the end of a horizontal spring (k = 120 N/m) is struck by a hammer, which gives it an initial speed of 2.76 m/s.

(a) Determine the period of the motion.
s
Determine the frequency of the motion.
Hz

(b) Determine the amplitude.
m

(c) Determine the maximum acceleration.
m/s²

(d) Determine the position as a function of time.
(  m ) sin[ (  rad/s)t ]

(e) Determine the total energy.
J

Hint- Pi is in radians

Answer 1

Given: k = 120 N/m, m = 755 g, v_i= 2.76 m/s

1) T = 2 x 3.14 sqrt(m/k) = 6.28 x 0.079 s = 0.498 s

2) w = 2 x 3.14 x f = 12.6 rad/s,

    f = 1/T = 2 Hz

3) Amplitude, A is given by,

kA²=m v_i²

A = 0.218 m

4) Maximum acceleration, a_max = -Aw² = -34.64 m/s²

5) The position, y = A sin (wt) = (0.218 m) sin (12.6t)

6) The total energy, E = K.E + P.E

Total energy is

K.E = (1/2) m (v_max)² = (1/2) x 0.755 x (wA)² = 2.85 J at x = 0

P.E = (1/2) k x² = (1/2) x 120 x A² = 2.85 J at x = A