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


T= 3.34
Vi= 0
A= 9.81
D= ?
d=Vit+1/2at^2
d= 1/2(9.81)(3.34)2
d= 54.7 or 55 meters tall
Answer:W = 1.23×10^-6BTU
Explanation: Work = Surface tension × (A1 - A2)
W= Surface tension × 3.142 ×(D1^2 - D2^2)
Where A1= Initial surface area
A2= final surface area
Given:
D1=0.5 inches , D2= 3 inches
D1= 0.5 × (1ft/12inches)
D1= 0.0417 ft
D2= 3 ×(1ft/12inches)
D2= 0.25ft
Surface tension = 0.005lb ft^-1
W = [(0.25)^2 - (0.0417)^2]
W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)
W = 1.23×10^-6BTU