Question

a mass weighing 10 newtons stretches a spring 10/49m. the mass is releases from 6 m below equilibrium with a downward velocity of 42 m/s in a median which offers a damping force numerically equal to 14 times the instantaneous velocity. use g=10m^2/s. (a) Find the equation of motion. y(t) = m. (b) find the time when the spring reaches its maximum displacement.

Answer 1

the weight of mass W = 10 N

@ equilibrium the extension of spring Xeq= 10/49 m

Let the spring constant be K

so we must have KXeq = W

K x 10/49 = 10

K = 49 N/m

Also let the co-ordinate of position be 0 @eqb.

PART A)

we have the mass released @ t = 0 @ y = - 6 m and V = dy/dt =  - 42 m/s

with g = 10 m/s2 as downwards is -y axis.

let after time t velocity be V m/s and extension be y(t) = y

so by newton's 2nd law

we have

W - Ky - bv = ma

a = dv/dt and m = 1kg

K = 49 and b = 14

10 - 49y - 14v = dv/dt

10 - 49y - 14dy/dt = d2y/dt2

we have

y'' + 14y' + 49y -10 = 0

which can be soved as

@ t = 0 y = -6

=> -6 = C1 + 10/49

=> C1 = -304/49

@ t = 0 dy/dt =  - 42

dy/dt =-7C1e-7t + C2( e-7t - 7t e-7t )

=> - 42 = -7C1+ C2

​​​=> C2 = 7C1 - 42

=> C2 = -304/7 - 42 = -598/7

PART B)

@ mas extension dy/dt =0

=> -7C1e-7t + C2( e-7t - 7t e-7t ) = 0

=>  C2( 1 - 7t ) = 7C1

=> 7t = 1 - 7C1/C2

t = 1/7*( 1 - 7C1/C2)

t = (1 - 304/598)/7 = 0.070235 s

-=-=-

Hope it helps!
Have a nice day!