Question
In: Physics
A 0.50-kg mass vibrates according to the equation x = 0.35 sin (5.50 t) where x...
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) the kinetic energy and potential energy when x = 0.3m Please Explain.
Solutions
Expert Solution
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
genius_generous answered 9 months agoRelated Solutions
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