Question

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

Answer 1

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

Step-by-step explanation:

Given the data in the question;

Using both Newton's and Hooke's law;

m
`x^{ff` + k
`x` = 0,
`x`(0) = 0,
`x^f`(0) = 1.5

given that mass m = 9 kg

`x` = 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
`x^{ff` + k
`x` = 0,

we input

9
`x^{ff` + 44.44
`x` = 0,

`x^{ff` + 4.9377
`x` = 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 )

⇒
`x^t`(t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )

using initial conditions

x(0) = 0

⇒ 0 = A

`x^t`(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 )