Question

A 442g mass vibrates according to the equation x = 0.370 sin (5.95 t) where x is in meters and t is in seconds.

A) Determine the amplitude.

B) Determine the frequency.

C) Determine the period.

D) Determine the total energy.

E) Determine the kinetic energy when x is 12.6 cm.

F) Determine the potential energy when x is 12.6cm.

Answer 1

x = 0.370 sin (5.95 t)

compare with x = A sin(wt)

a)
Amplitude A=0.370 m

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b)
frequency , f = w/2pi = 5.95/2pi = 0.947 Hz

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c)

period , T = 1/f = 1.05 sec

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d)

total energy = 0.5KA^2 = 0.5*mw^2 A^2 = 0.5*0.442*5.95^2*0.370^2

E = 1.07 J

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e)

v = w *sqrt(A^2-x^2) = 5.95*sqrt(0.37^2 - 0.126^2)

v = 2.07 m/s

Kinetic energy = 0.5mv^2 = 0.5*0.442*2.07^2 = 0.9468 J

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f)

potetial energy

U = E-K = 1.07-0.9468 = 0.123 J

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