need help, step by step please 1. Consider an oscillator with mass 0.343 kg and

Question

need help, step by step please

1. Consider an oscillator with mass 0.343 kg and spring constant 12.8 N/m. If the amplitude of the oscillation is 6.99 cm, what is the maximum speed?(Note, you might already have solved similar problems employing conservation of energy. Here, I'd like you to find the answer using the equations of motion.)Answer in m/s.

2. Consider an oscillator with frequency 17 Hz. If at t=0 the oscillator is at a maximum displacement from the equilibrium of +3.05 cm, what is the displacement 3.45 seconds later? Answer in cm with the proper sign.

3. Consider an oscillator with mass 0.3 kg and spring constant 18.7 N/m. 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) maximum speed is Vmax = A*w = 0.0699*w

w = sqrt(k/m) = sqrt(12.8/0.343) = 6.1 rad/s

then Vmax = 0.0699*6.1 = 0.427 m/s

2) let = x =A*cos(wt)

then at t = 3.45 s later

x = 0.0305*cos(2*pi*f*t) = 0.0305*cos(2*3.142*17*3.45) = -0.0167 m = -1.67 cm

3) m = 0.3 kg

k = 18.7 N/m

A = 6.02 cm

then let postion after t seconds is x = A*cos(w*t)

w = sqrt(k/m) = sqrt(18.7/0.3) = 7.89 rad/s

t = ?

x = -3.41 cm

then -3.41 = 6.02*cos(7.89*t)

-0.566 = cos(7.89*t)

7.89*t = acos(-0.566) = 2.17

t = 2.17/7.89 = 3.63 S

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