# An object with mass 3.0 kg is attached to to a spring with spring stiffness constant k = 280 N/m and is?

executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of speed of 0.55 m/s (a) calculate the amplitude of the motion. (b) calculate the maximum velocity attained by the object?.

### 2 Answers

- anil bakshiLv 71 decade agoBest Answer
for spring restorative force

F = - k x

m dv / dt = - k x

dv = - k/m x dt

dv = - k/m x {dx/v}

v dv = - k/m dx integrating

(v^2/2) = - k/m (x^2/2) + c

(v^2) = - k/m (x^2) + c1

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at x = a (amplitude)

v = 0 (spring retreats after reaching amplitude

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0 = - k/m (a^2) + c1 >>> c1 put back

(v^2) = - k/m (x^2) + k/m (a^2)

v^2 = k/m [a^2 - x^2]

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given x = 0.02, v = 0.55

0.55^2 = k/m [a^2 - 0.02^2]

a^2 = [(m/k) * 0.55^2] + 0.02^2

calculate (a) amplitude

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v^2 = k/m [a^2 - x^2]

velocity will be maximum when x = 0 ie at mean position

by seeing v^2 formula you can know that (- x^2) term is trying to decrease the v^2 so if x=0 this term will not reduce v^2

hence vmax

(vmax)^2 = k/m * a^2

just found value of a is known and others are known

vmax = a √(k/m)

Source(s): hope you will settle these questions - kaseeLv 43 years ago
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