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Question

Acceleration of 3rd maxima w.r.t 3rd maxima on other side of central maxima at t=3s is

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A
0.02ms2^i
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B
0.03ms2^i
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C
10ms2^i
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D
0.6ms2^i
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Solution

The correct option is C 0.03ms2^i
Position of 3rd maxima, y=3λDd
Differentiating twice, a3=2λdd2dt2(D)
As D=1+12gt2
Thus a3=3λgd=.015ms2^i (for right side of central maxima)
Similarly a3=3λgd=.015ms2^i (for left side of central maxima)
Now acceleration of 3rd maxima w.r.t 3rd maxima on other side ofcentral maxima, a=.015ms2^i(.015)ms2^ia=.03ms2^i

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