A particle with mass 1.59 kg oscillates horizontally at the end of a horizontal

Question

A particle with mass 1.59 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.997 m and a duration of 129 s for 6.0 × 101 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 63.9% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position.

Explanation / Answer

A)

A = 0.997 m

T = time period = 129 / 60 = 2.15 s = 2*pi*sqrt(m/k)

k = spring constant = 4 * (pi)^2 * m / (2.15)^2 = 13.58 N/m

frequency = 1/T = 1/2.15 = 0.465 Hz

potential energy = kA^2/2 = 13.58 * 0.997^2/2 = 6.75 J

for max speed (energy balance)

m(v_max)^2/2 = kA^2/2 = 6.75

V_max = sqrt(2*6.75 / 1.59) = 2.91 m/s

B)

x= 0.639*A = 0.639*0.997 = 0.637 m

PE = kx^2/2 = 13.58 * 0.637^2 / 2 = 2.76J

PE + KE = 6.75 from energy conservation

2.76 + KE = 6.75

KE=mv^2/2 = 3.99J

v = 2.24 m/s

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