Help! Xeq = m A = m upsilon max = m/s amax = m / s2 k = N/m Etot = j (Scroll dow

Question

Help!

Xeq = m A = m upsilon max = m/s amax = m / s2 k = N/m Etot = j (Scroll down for more answer blanks.) A 0.237 - kg particle undergoes simple harmonic motion along the horizontal x - axis between the points x1 = -0.309 m and x2 = 0.473 m. The period of oscillation is 0.635 s. Find the frequency, f, the equilibrium position, Xeq, the amplitude, A, the maximum speed, Vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot. F = Hz Xeq = m A = m upsilon max = m/s (Scroll down for more answer blanks.)

Explanation / Answer

a) f = 1/T = 1/.635=1.575 Hz
b) xeq is mid point between x1 and x2
xeq = (-0.309+0.473)/2=0.082 m
C) A = x2 - xeq = 0.473-0.077=0.396 m
first lets do G)
E = 1/2 m w^2 A^2 = 0.5*0.237*(2*pi*1.575)^2*(0.396)^2=1.820 J
D)
E = 1/2 mv^2
1.820 = 0.5*0.237*v^2
v=3.92 m/s
E) a max = w^2 A = (2*pi*1.575)^2*0.396=38.78
F) w = sqrt(k/m)
k = m w^2 = 0.237*(2*pi*1.575)^2=23.21

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