Analyzing Newton\'s 2nd Law for a mass spring system, we found ax = - k/m x . Co

Question

Analyzing Newton's 2nd Law for a mass spring system, we found ax = - k/m x . Comparing this to the x-component of uniform circular motion, we found as a possible solution for the above equation: x = Acos(wt) vx = - wAsin(wt) ax = - w^22Acos(cut) with W = squreroot k/m. and A the amplitude (maximum displacement from equilibrium). Consider an oscillator with mass 0.217 kg and spring constant 21.6 N/m. If the amplitude of the oscillation is 4.53 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. Consider an oscillator with frequency 33 Hz. If at t = 0 the oscillator is at a maximum displacement from the equilibrium of + 4.78 cm, what is the displacement 3.85 seconds later? v Question Completion Status: Answer in cm with the proper sign. Consider an oscillator with mass 0.31 kg and spring constant 16.1 N/kg. The oscillator is displaced from the equilibrium position by + 5.31 cm and released. After what time, is the oscillator for the first time at a displacement of -2.07 cm? Click Save and Submit to save and submit. Click Save All Answers to save all answers.

Explanation / Answer

Here ,

mass , m = 0.217 Kg

spring constant , k = 21.6 N/m

ampliude , A = 4.53 cm

let the maximum speed is v

Using conseravtion of energy

0.5 * k * A^2 = 0.5 * m * v^2

21.6 * 0.0453^2 = 0.217 * v^2

solving for v

v = 0.452 m/s

the maximum speed of the block is 0.452 m/s

b)

for the oscilator ,

y = 4.78 * cos(2*pi*f*t)

y = 4.78 * cos(2 * pi * 33 * 3.85)

y = 4.54 cm

the position of the particle is 4.54 cm

3)

as w = sqrt(k/m)

w = sqrt(16.1/.31)

w = 7.21 rad/s

y = A * cos(w*t)

-2.07 = 5.31 * cos(7.21 * t)

t = 0.273 s

the time is 0.273 s

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