A spring hangs with its upper end fixed. An object with mass 50 g at its lower end stretches it 9.8 cm. Then this object is replaced by another whose mass is 5 kilograms and once it reaches its equilibrium position, the object is carried up 0.26 m and released without momentum. The object receives an external force equivalent to 12cos (2t) dynes, and there are no damping forces. If the positive direction is taken downwards, then a differential equation and initial conditions that allow determining the position x (t), in cm, of the object at t seconds, is: If necessary, use the constant of gravity g as g = 980 cm / s2

icon
Related questions
Question

A spring hangs with its upper end fixed. An object with mass 50 g at its lower end stretches it 9.8 cm. Then this object is replaced by another whose mass is 5 kilograms and once it reaches its equilibrium position, the object is carried up 0.26 m and released without momentum. The object receives an external force equivalent to 12cos (2t) dynes, and there are no damping forces. If the positive direction is taken downwards, then a differential equation and initial conditions that allow determining the position x (t), in cm, of the object at t seconds, is:

If necessary, use the constant of gravity g as g = 980 cm / s2

the answers are in the attached image.

 

a) 5000z" + 5000x = 12 cos(2t), ¤(0) = –26, x'(0) = 0
b) 500z" + 5000x = 12 cos(2t), x(0) = -26, z'(0) = 0
c) 5000z" + 5000x = 12 cos(2t), æ(0) = 26, x'(0) = 0
d) 50z" + 9.8x = 12 cos(2t), ¤(0) = 26, x'(0) = 0
Transcribed Image Text:a) 5000z" + 5000x = 12 cos(2t), ¤(0) = –26, x'(0) = 0 b) 500z" + 5000x = 12 cos(2t), x(0) = -26, z'(0) = 0 c) 5000z" + 5000x = 12 cos(2t), æ(0) = 26, x'(0) = 0 d) 50z" + 9.8x = 12 cos(2t), ¤(0) = 26, x'(0) = 0
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer