A frictionless spring with a 9-kg mass can be held stretched 1.8 meters beyond its natural length by a force of 80 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after tt seconds. meters

the required solution is; x(t) = 0.675sin( 2.222t )

Explanation:

Given the data in the question;

Using both Newton's and Hooke's law;

m + k = 0, (0) = 0, (0) = 1.5

given that mass m = 9 kg

= 1.8 m

k is F / x

hence

k = F / x

given that, F = 80 N

we substitute

k = 80 / 1.8

k = 44.44

so

m + k = 0,

we input

9 + 44.44 = 0,

+ 4.9377 = 0

so auxiliary equation is,

r² + 4.9377 = 0

r² = -4.9377

r = √-4.9377

r = ±2.222i

hence, the solution will  be;

x(t) = A×cos( 2.222t ) + B×sin( 2.222t )

⇒ (t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )

using initial conditions

x(0) = 0

⇒ 0 = A

(t) = 1.5

1.5 = 2.222B

so

B = 1.5 / 2.222 = 0.675

Hence, the required solution is; x(t) = 0.675sin( 2.222t )

no. dot product   =0 when vecs at right angles

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.