Answer:
u" + 40u' + 49u = 2 sin(t/6)
upp + 40up + 49u = 2 sin(t/6)
Explanation:
Step 1: Data given
mass = 5 kg
L = 20 cm = 0.2 m
F = 10 sin(t/6)N
Fd(t) = - 6 N
u(0) = 0.03 m/s
u(0) = 0
u'(0) = 3 cm/s
Step 2:
ω =kL
k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²
Since Fd(t) = -γu'(t) we know:
γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m
The initial value problem which describes the motion of the mass is given by
5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03
This is equivalent to:
u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03
upp + 40up + 49u = 2 sin(t/6)
With u in m and t in s
Answer:4.93 m/s
Explanation:
Given
height to reach is (h )1.24 m
here Let initial velocity is u
using equation of motion
here Final Velocity v=0
a=acceleration due to gravity
u=4.93 m/s
They are incline hope this helps!
False. p waves, than s waves
