A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motion is a. `0^@, b. 60^@, c. 90^@`

Answer

(a) `7 cm` (b) `sqrt(37) cm = 6.1 cm` (c) `5 cm`

Explanation

`A_(r) = sqrt(A_(1)^(2)+A_(2)^(3)+2A_(1)A_(2)costheta)`
(a) `A_(r) = sqrt(3^(2)+4^(2)+2xx3xx4xxcostheta) = 7 cm` (b) `A_(r) = sqrt(3^(2)+4^(2)+2xx3xx4xxcos60) = sqrt(37)cm = 6.1cm`
(c) `A_(r) = sqrt(3^(2)+4^(2)+2xx3xx4xxcos90) = 5 cm`