A particle with mass 125 kg oscillates horizontally at the e

A particle with mass 1.25 kg oscillates horizontally at the end of a horizontal spring. A student measures Map amplitude of 0.881 m and a duration of 131 s for 72 cycles of oscillation. Find the frequency, f, the speed a the equilibrium position, Mmax, the spring constant, k, the potential energy at an endpoint, Unax, the potential energy when the particle is located 46.5% of the amplitude away from the equilibrium position, U, and the kinetic energy, K, and the speed, v, at the same postion. Number f0.55 Hz Number m/s max Number N/ m Number (Scroll down for more answer blanks.) max Number U= 11

Solution

According to the given problem,

T = 131 / 72 = 1.82 s
a) f = 1 / T

f = 78 / 121 = 0.55 Hz

c) T = 2 (m / k)

131 / 72 = 2 (1.25 / k)

k = 14.90 N/m

d) U_max = k x² / 2

U_max = 14.90*0.881² / 2

U_max = 5.78 J

b) U_max = K.E_max = m*v²/2

5.78 = 1.25 v² / 2

V_max = 3.04 m/s

46.5% of the amplitude away from the equilibrium position is

0.881*0.465 = 0.5m

At that point the elastic potential energy is:

e) U = k x² / 2

U = 14.90 *0.5² / 2

U = 1.863J

f) At that point the kinetic energy is:

K = U_max - U
K = 5.78 - 1.863
K = 3.917 J

g) At that point the velocity is:

K = m *v² / 2

3.917 = 1.25* v² / 2

v = 2.50m/s