A particle with mass 1.53 kg oscillates horizontally at the end of a horizontal

Question

A particle with mass 1.53 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.941 m and a duration of 123 s for 68 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 32.5% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position.

f = ___ Hz

vmax = ___ m/s

k = ___ N/m

Umax = ___ J

U = ___ J

K = ___ J

v = ___ m/s

Explanation / Answer

T = 123 / 68 = 1.808 s

f = 1 / T
f = 68 / 123 = 0.552 Hz < - - - - - answer

T = 2 (m / k)
123 / 68 = 2 (1.53 / k)
k = 18.478 N/m < - - - - - answer

Umax = k x² / 2
Umax = 18.478 0.941² / 2
Umax = 8.18 J < - - - - - answer

Umax = Ekmax = m v² / 2
8.18 = 1.53 v² / 2
Vmax = 3.269 m/s < - - - - - answer

32.5% of the amplitude away from the equilibrium position is
0.941 0.325 = 0.3058 m

At that point the elastic potential energy is:
U = k x² / 2
U = 18.478 0.3058² / 2
U = 0.8639J < - - - - - answer

At that point the kinetic energy is:
K = Umax - U
K = 8.18 - 0.8639
K = 7.3161 J < - - - - - answer

At that point the velocity is:
K = m v² / 2
7.3161 = 1.53 v² / 2
v = 3.092 m/s < - - - - - answer

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