Answer:
The number of turns of the inductor is 2000 turns.
Explanation:
Given;
emf of the inductor, E = 0.8 V
the rate of change of current with time, dI/dt = 10 A/s
steady current in the solenoid, I = 0.2 A
flux per turn, Ф = 8.0 μWb per
Determine the inductance of the solenoid, L
E = L(dI/dt)
L = E / (dI/dt)
L = 0.8 / (10)
L = 0.08 H
The inductance of the solenoid is given by;

Also, the magnetic field of the solenoid is given by;

I is 0.2 A



But Ф = BA

Therefore, the number of turns of the inductor is 2000 turns.
Answer: y(t)= 1/π^2 sin(6*π^2*t)
Explanation: In order to solve this problem we have to consider the general expression for a harmonic movement given by:
y(t)= A*sin (ω*t +φo) where ω is the angular frequency. A is the amplitude.
The data are: ν= 3π; y(t=0)=0 and y'(0)=6.
Firstly we know that 2πν=ω then ω=6*π^2
Then, we have y(0)=0=A*sin (6*π^2*0+φo)= A sin (φo)=0 then φo=0
Besides y'(t)=6*π^2*A*cos (6*π^2*t)
y'(0)=6=6*π^2*A*cos (6*π^2*0)
6=6*π^2*A then A= 1/π^2
Finally the equation is:
y(t)= 1/π^2 sin(6*π^2*t)