Use the worked example above to help you solve this problem. (a) Find the amplit

Question

Use the worked example above to help you solve this problem. (a) Find the amplitude, frequency, and period of motion for an object vibrating at the end of a horizontal spring if the equation for its position as a function of time is the following. x = (0.205 m) cos (pi/8.00 t) A = m f = Hz T = s (b) Find the maximum magnitude of the velocity and acceleration. v_max = m/s a_max = m/s^2 (c) What are the position, velocity, and acceleration of the object after 1.05 s has elapsed? x = m v = m/s a = m/s^2 If the object-spring system is described by x = (0.310 m) cos (1.45 t), find the following. (a) the amplitude, the angular frequency, the frequency, and the period A = m omega = rad/s f = Hz T = s (b) the maximum magnitudes of the velocity and the acceleration v_max = m/s a_max = m/s^2 (c) the position, velocity, and acceleration when t = 0.250 s x = m v = m/s a = m/s^2

Explanation / Answer

PRACTICE IT:

(A) A = 0.205 m

f = (pi / 8) / 2pi = 0.0625 Hz

T = 1/f = 16 s

(B) Vmax = A w = 0.205 x pi / 8 = 0.0805 m/s

a_max = A w^2 = 0.0316 m/s^2

(C) cos(pi x 1.05 / 8) = 0.916

sin(pi x 1.05 / 8) = 0.40

x = 0.205 x 0.916 = 0.188 m

v = 0.0805 x 0.40 = -0.0322 m/s

a = 0.0316 x 0.916 = - 0.0289 m/s

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exercise:

A) A = 0.310 m

w = 1.45 rad/s

f = w/ 2pi = 0.23 Hz

T = 1/f = 4.33 s

(B) Vmax = A w = 0.310 x 1.45 = 0.45 m/s

a_max = A w^2 = 0.65 m/s^2

(C) cos(1.45 x 0.250) = 0.935

sin(1.45 x 0.250 ) = 0.355

x = A cos(wt) = 0.29 m

v = - Vmax sin(wt) = - 0.16 m/s

a = -a_max cos(wt) = - 0.609 m/s^2

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