## 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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