a force of 400 newtons stretches a spring 2 m. a mass of 50 kg is attached to the end of the spring and released from the equilibrium position with an upward velocity of 10 m/s. find the equation of motion.



Answer :

The equation of motion is [tex]x_{(t)}[/tex] = −5 sin 2t

What is motion?

Motion, a phenomenon in which an object tends to change its position with respect to time. Motion is mathematically can be described in terms of distance, displacement, speed, velocity, acceleration etc.

For the given case;

Let m be the mass and k be the spring constant. The Newton's second law for the system states:

[tex]m\frac{d^{2}x}{dt}[/tex] = -kx

where [tex]x_{(t)}[/tex] is the displacement from the equilibrium position. The equation can be transformed into

[tex]\frac{d^{2}x}{dt}[/tex] + [tex]\frac{k}{m}x[/tex] = 0

From Hooke's law we can calculate the spring constant k

k = [tex]\frac{W}{s}[/tex] Where, W =400N and s = 2 m

[tex]\frac{400}{2}[/tex] = 200 N/m

Substitute the values in above equation:

m = 50 kg and k = 200 N/m

So,

[tex]\frac{d^{2}x}{dt}[/tex] + 4x = 0

Since, xⁿ + ω²x = 0

[tex]x_{(t)}[/tex] = c₁ cos 2t + c₂ sin 2t

[tex]x_{(0)}[/tex] = 0 ft and [tex]x'_{(0)}[/tex] = −10 m/s

The final solution will be [tex]x_{(t)}[/tex] = −5 sin 2t

To know more about motion visit:

https://brainly.com/question/21140538

#SPJ4

Other Questions