A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.8 cm. (a) Calculate the maximum value of its speed. (b) Calculate the maximum value of its acceleration. (c) Calculate the value of its speed when the object is 8.80 cm from the equilibrium position. cm/s (d) Calculate the value of its acceleration when the object is 8.80 cm from the equilibrium position. cm/s2 (e) Calculate the time interval required for the object to move from x = 0 to x = 6.80 cm.

for A i got 50.2057 cm/s

B 196.89 cm/s2

those are right i just can't get the rest

Explanation / Answer

m = 0.52 (kg)
k = 8 (N/m)
A = 12.8 (cm) = 1.28e-1 (m)
w = sqrt(k/m) = sqrt(16) = 3.92 (rad/s)

a) When x = 6.8 (cm) = 0.68e-1 (m):
F = -k*x = -8*0.68e-1 = -5.44e-1 = -0.544 (N)

Since x(t) = 1.28e-1 * sin(3.92*t),
x'(t) = 5.02e-1 * cos(3.92*t)

0.68 = sin(3.92*t) => cos(3.92*t) = sqrt(1 - sin^2(3.92t))
= sqrt(1 - 0.68^2)
= 0.73
So x'(t) = 5.02e-1 * 0.73 = 3.6e-1 = 0.36 (m/s)

b) 0 = x when t = 0;
6.8 = x(t) = 12.8 * sin(3.92t) when:
0.53 = sin(3.92t)
=> t = Arcsin(0.53) / 3.92 = 0.21 (s)

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