Harmonic Please! show me how to do this correctly! show units and work Comparing
ID: 1467091 • Letter: H
Question
Harmonic Please! show me how to do this correctly! show units and work
Comparing this to the x-component of uniform circular motion, we found as a possible solution for the above equation: x = A cos(omega t) v_x = - omega A sin(omega t) a_x = - omega^2 A cos(omega t) with omega = square root k/m, and A the amplitude (maximum displacement from equilibrium). Consider an oscillator with mass 0.365 kg and spring constant 18 N/m. If the amplitude of the oscillation is 4.07 cm. what is the maximum speed? Answer in m/s. Consider an oscillator with frequency 28 Hz. If at t=0 the oscillator is at a maximum displacement from the equilibrium of +5.88 cm, what is the displacement 3.88 seconds later? Answer in cm with the proper sign. Consider an oscillator with mass 0.3 kg and spring constant 18.7 N/kg. The oscillator is displaced from the equilibrium position by +6.02 cm and released. After what time, is the oscillator for the first time at a displacement of -3.41 cm?Explanation / Answer
1)) angular frequency, w = sqrt(k/m)
= sqrt(18/0.365)
= 7.02 rad/s
Aplitude, A = 4.07 cm
we know, Vmax = A*w
= 4.07*7.02
= 28.6 cm/s
= 0.286 m/s <<<<<<<<<-----------Answer
2) angu;ar frequency, w = 2*pi*f
= 2*pi*28
= 175.84 rad/s
A = 5.88 cm
so, x = A*cos(w*t)
at time t = 3.88s
x = 5.88*cos(175.84*3.88)
= -5.06 cm <<<<<<<<<-----------Answer
3) angular frequency, w = sqrt(k/m)
= sqrt(18.7/0.3)
= 7.9 rad/s
A = 6.02 cm
Again, Apply, x = A*cos(w*t)
-3.41 = 6.02*cos(7.9*t)
==> cos(7.9*t) = -3.41/6.02
cos(7.9*t) = -0.566
7.9*t = cos^-1(-0.566)
7.9*t = 2.17
t = 2.17/7.9
= 0.275 s <<<<<<<<<-----------Answer
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