A particle moves with simple harmonic motion in a straight line. In first tau s, after starting from rest it travels a distance a, and in next tau s it travels 2a, in same direction, then :

A particle moves with simple harmonic motion in a straight line. In fi

1.	A particle moves with simple harmonic motion in a straight line. In first tau s, after starting from rest it travels a distance a, and in next tau s it travels 2a, in same direction, then :
Answer» time period of oscillations is `6tau` amplitude of motion is 3a time period of oscillations is `8tau` amplitude of motion is 4a. Solution :`A(1-cos OMEGA tau)=a,A(1-cos 2OMEGA tau)=3a` `cos omega tau=(1-(a)/(A)),cos 2 omega tau=(1-(3a)/(A))` `2(1-(a)/(A))^(2)-1=1-(3a)/(A)` Solving the EQUATION `(a)/(A)=(1)/(2)impliesA=2a` `cos omega tau=(1)/(2),T=6tau` So correct choice is (a).

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A particle moves with simple harmonic motion in a straight line. In first tau s, after starting from rest it travels a distance a, and in next tau s it travels 2a, in same direction, then :

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