HC Verma - Concepts of Physics Part 1

Book: HC Verma - Concepts of Physics Part 1

Chapter: 12. Simple Harmonic Motion

Subject: Physics - Class 11th

Q. No. 55 of Exercises

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55

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


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