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Ask Your Teacher A force of 640 newtons stretches a spring 4 meters. A mass of 40 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 8 m/s. Find the equation of motion.

Answer :

aabdulsamir

Answer: -(4 sin2t)

Explanation:

Let m = mass attached

k = spring constant

Using newton's second law of motion,

Md²x/dt² = -kx

where x(t) = displacement from equilibrium position.

d²x/dt² + (k/m)*x = 0

Using Hooke's law,

W = Ks

K = w/s = 640/4 = 160N/m

Put m = 40kg and k = 160N/m

d²x/dt² + (160/40)x = 0

d²x/dt² + 4x = 0

Note: w = k/m = 160/40 = 4 rad/s

x¹¹ + ω²x = 0

the general solution is,

x(t) = c₁ cos2t + c₂sin2t

Using the conditions

x(0) = 0ft

x¹(0) = -8m/s

x(t) = -4sin2t

akande212

Answer:

y = 4cos(2t) assuming phase angle = 0°

Explanation:

Given that a force of 640N stretches a spring 4m

k = 640/4 = 160N/m

m = mass attached = 40kg

ω = √(k/m) = √(160/40) = 2rad/s

The amplitude of the simple harmonic motion is given by

A = √(yo² + voy²/ω²)

yo = 0m

voy = initial upward speed. = 8m/s

So

A = √(0² + 8²/2²) = √(64/4) = 4m

So y = Acos(ωt) = 4cos(2t)

y = 4cos(2t) assuming phase angle = 0°

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