Three simple harmonic motions in the same direction having each of amplitude "a" and the same period are superposed. If each differs in phase from the next by `pi//4` then

Options

A. Resultant amplitude is `(sqrt(2)+1)` a
B. Phase of resultant motion relative to first is `90^(@)`
C. The energy associated with the resulting motion is `(3+2 sqrt(2))` time the energy associated with any single motion
D. Maximum speed of resultant SHM will be more than double of the initial SHM's

Answer

A, C, D

Explanation

(A)The vector sum of these three is `(sqrt(2)+1)` a
(C) Energy associated with each wave `=Ka^(2)`
Energy with resultant motion `=K(sqrt(2)+1)^(2)a^(2) Rightarrow (3+2sqrt(2))`
(D) `V_(max)=Aomega`
`V_(initial)=aw`
`V_(resultant max =(sqrt(2)+1) a omega=2.14a omega`
:.`V_(resultant) 72 time V_(initial)`