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Explanation :
It is given that,
Diameter of the coil, d = 20 cm = 0.2 m
Radius of the coil, r = 0.1 m
Number of turns, N = 3000
Induced EMF, 
Magnitude of Earth's field, 
We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :




So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.
We can solve the problem by using the first law of thermodynamics:

where
is the change in internal energy of the system
is the heat absorbed by the system
is the work done by the system on the surrounding
In this problem, the work done by the system is

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

with a negative sign as well because it is released by the system.
Therefore, by using the initial equation, we find

As we know that KE and PE is same at a given position
so we will have as a function of position given as

also the PE is given as function of position as

now it is given that
KE = PE
now we will have




so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula

now to find the fraction of kinetic energy



now since total energy is sum of KE and PE
so fraction of PE at the same position will be

