A 5.00 kg object on a horizontal frictionless surface is attached to a spring wi

Question

A 5.00 kg object on a horizontal frictionless surface is attached to a spring with k = 1000 N/m. The object is displaced from equilibrium 50.0 cm horizontally and given an initial velocity of 10.0 m/s back toward the equilibrium position. What are the motion's frequency, the initial potential energy of the block-spring system, the initial kinetic energy, and the motion's amplitude? Figure 15-38 shows the kinetic energy K of a simple harmonic oscillator versus its position x. The vertical axis scale is set by K_s = 4.0 J. What is the spring constant? A 10 g particle undergoes SHM with an amplitude of 2.0 mm, a maximum acceleration of magnitude 8.0 Times 10^3

Explanation / Answer

(a)

motion frequency f = (1/2pi)*sqrt(k/m)

k = spring constant

m = mass attached

f = (1/(2pi))*sqrt(1000/5)

f = 2.25 Hz

+++++++++++++++

(b)

initial potential energy Ui = 0.5*k*x^2 = 0.5*1000*0.5^2 = 125 J

(c)

initial kinetic energy ki = 0.5*m*v^2 = 0.5*5*10^2 = 250 J

(d)

total energy E = Ui + ki = 375 J

but toal energy E = (1/2)*K*A^2

(1/2)*1000*A^2 = 375

A = 0.866 m <<<--------answer

+++++++++++++++++

Kmax = 0.5*m*vmax^2

spring constant k = m*w^2

Vmax = A*w

Kmax = 0.5*m*w^2*A^2

Kmax = 0.5*k*A^2

from the graph A = 0.12 m

Kmax = 6

0.5*k*0.12^2 = 6

k = 833.33 N/m

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