Answer:
1.63 
Explanation:
The equation 
would be used where:
g = acceleration due to gravity
G = universal gravitational constant (6.67408 × 
)
M = mass
R = radius.
assuming the values following both 10s of the mass and radius are their exponents, which is a little confusing, the values given would simply be substituted.
Anyone can correct me if I'm wrong.
Answer:
equation of motion for the mass is x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )
Explanation:
Given data
mass = 3 slugs = 3 * 32.14 = 96.52 lbs
constant k = 9 lbs/ft
Beta = 6lbs * s/ft
mass is pulled = 1 ft below
to find out
equation of motion for the mass
solution
we know that The mass is pulled 1 ft below so
we will apply here differential equation of free motion i.e
dx²/dt² + 2 α dx/dt + ω² x =0 ........................1
here 2 α = Beta / mass
so 2 α = 6 / 96.52
α = 0.031
α² = 0.000961 ...............2
and
ω² = k/mass
ω² = 9 /96.52
ω² = 0.093 ..................3
we can say that from equation 2 and 3 that α² - ω² = -0.092239
this is less than zero
so differential equation is
x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )
equation of motion for the mass is x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )
Answer:
the time between successive meridian transits of the sum at a particular place
