Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The displacement of a particle varies according to the relation X = 4 (cos pi t+sin pi t). The amplitude of the particle is :

Paragraph : A particle vibrates in S.H.M. along a straight line. Its v

1.	Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The displacement of a particle varies according to the relation X = 4 (cos pi t+sin pi t). The amplitude of the particle is :
Answer» `-4` 4 `4sqrt(2)` 8 Solution :`X=4 COS pi t+4sin pi t` since `cos pi t` and `sin pi t` are out of phase by `90^(@)`. `:.` Amplitude `A=sqrt(4^(2)+4^(2))=4sqrt(2)` Hence CORRECT choice is ( C ).

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Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The displacement of a particle varies according to the relation X = 4 (cos pi t+sin pi t). The amplitude of the particle is :

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