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A 0.50-kg mass vibrates according to the equation x = 0.35 sin (5.50 t) where x...

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Physics

A 0.50-kg mass vibrates according to the equation x = 0.35 sin(5.50 t) where x is in meters and t is in seconds. Determine (A)the amplitude, (B) the frequency, (C) the total energy, and (D) thekinetic energy and potential energy when x = 0.3m PleaseExplain.

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3.7 Ratings (367 Votes)

Given equation X= 0.35 sin(5.50 t), mass m = 0.50 kg

comparing the equation with X= A Sin (wt), A is amplitude, w is angular frequency = 2*pi*f, f is frequency

A) Amplitude A = 0.35 m

B) frequency f = w/2pi = 5.50 / (2*3.14) = 0.8758 Hz

C) total energy E = 1/2 m w^2A^2 = 0.5 *0.5*5.5^2*0.35^2 = 0.9264 J

D) potential energy P.E = 1/2 kx^2                              w = sqrt(k/m) ==> k= w^2m = 5.5^2*0.5 = 15.125

                                  = 0.5*15.125*0.3^2                

                                 = 0.6806 J

total energy = kinetic energy + potential energy

kinetic energy = total energy - potential energy

                     = 0.9264 -0.6806

               K.E = 0.2458 J
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