The electrical resistance of the wire is proportional to its length L and inversely proportional to the square of its diameter,

.
The length of the wire in the problem is changed from 200 inches to 500 inches, so the new length is 2.5 times the initial length:

the diameter of the wire is not changed, so the new electrical resistance must be 2.5 times the original value:
The displacement is x + (60km - 45km) - x =60km -45 km = 15 km
U can infer that it’s a graph line
Period is T = 1/f. The frequency, f, is 5cycles/10s = 0.5. So the period T=1/0.2 = 2.