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 motions is (a) 0°, (b) 60o, (c) 90o.
(a) 7.0 cm
(b) 6.1 cm
(c) 5.0 cm
a) Given amplitudes of S.H.Ms
= 3.0 cm
= 4.0cm
Given that both S.H.Ms have same time period and direction and hence their equations will be
Sinωt
Sinωt
Resultant S.H.M equation will be y = = Sinωt + Sinωt
= 3Sinωt + 4Sinωt = 7sinωt
Amplitude of resultant S.H.M is 7cm
b) Here the S.H.Ms have a phase difference of π/3 between them
And hence Sinωt
Sinωt+ π/3
Resultant amplitude will be A=
= = = 6.1 cm
c) Here the phase difference is π/2
Sinωt
Sinωt+ π/2
Resultant amplitude will be A=
=
= 5cm