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A particle is simultaneously subjected to three SHM, all the same frequency and in the same direction. If the amplitude are 0.5 mm , 0.4 mm and 0.3 mm respectively and the phase difference between 1st and 2nd is 45° and between the second and the third is 30° , then

(a)  The amplitude of resultant between 1st and 2nd motion is 0.5 mm

(b)  The phase difference of the resultant of 1st and 2nd with 1st motion is 20°

(c)  The amplitude of the resultant displacement is 1.03 mm

(d)  The phase of the resultant motion relative to first motion is 33.7°

1 Answer

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Best answer

Correct option (b), (c) and (d) 

Explanation:

Here  A= 0.5 mm, A = 0.4 mm, phase difference between them is  φ = φ2 - φ1 = 45° The resultant amplitude A is

Thus option (a) is not correct

Phase difference of the resultant OP1 and P1P2 relative to the 1st component OP1 is α In ΔOP1P2 we have

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