The equation of motion of mass attached to a spring is x =2.95 cos 3.32t, where

Question

The equation of motion of mass attached to a spring is x =2.95 cos 3.32t, where x is in meters and t is in seconds. The mass is 0.58 kg. Find the (a) amplitude, (b) frequency, (c) total energy, (d) potential energy and kinetic energy when x = 1.57 meters

I know the answer is:

a) I do not know

b) f = 0.528 Hz

c) k = 6.39 N/m , Etot = 27.8 J

d) PE = 7.88 J , KE = 19.9 J

but I do not know how to do the problem. This is a trigonometry based physics course so please do not use calculus in your answer. Please show all work. Thank you

Explanation / Answer

given eq is

x = 2.95*cos(3.32*t)

so amplitude (A) = 2.95 m

angular frequency (w) = 3.32 rad/s

frequency = w/(2*pi) = 3.32/(2*22/7) = 0.528 Hz

KE = (1/2)mv^2(t) = (1/2)mw^2*A^2*(sin(3.32t))^2

and k = w^2*m = 3.32^2*0.58 = 6.39 N/m

KE = (1/2)*kA^2*(sin(3.32t))^2

PE = (1/2)kx^2(t) = (1/2)*kA^2*(cos(3.32t))^2

E(total) = KE + PE = (1/2)kA^2 = (1/2)*6.39*(2.95)^2 = 27.8 J

x = 1.57 m so

1.57 = 2.95*cos(3.32t)

0.533 = cos(3.32t)

1.0095953 = 3.32*t

t = 0.304 sec

KE = (1/2)*kA^2*(sin(3.32t))^2 = (1/2)*6.39*(2.95)^2*(0.84661653)^2 = 19.92 J

PE = (1/2)*kA^2*(cos(3.32t))^2 = (1/2)*6.39*(2.95)^2*(0.533)^2 = 7.876 J

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