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Q.

A particle starts its SHM on a line at initial phase of π/6.It reaches again the point of start after time t. lt crosses yet another point P on the same line at successive intervals 2t and 3t respectively. Find the amplitude of the motion, if the particle crosses the point of start at speed 2 m/s:

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a

8tπ

b

4tπ

c

2tπ

d

tπ

answer is A.

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Detailed Solution

x=Asin⁡ω×0+π6=Asin⁡π6=A2 (t=0 at initial phase )Let time t=0 is considered from point Q where displacement =A/2 Time period T=tQA+tAQ+tQP+tPQ+tPB+tBP=t+[(2t−t)+(2t−t)]+t=4t=2π/ωω=π/2t;x=Asin⁡ωt+π6dxdt=Aωcos⁡ωt+π6=Aωcos⁡π2+π62=A×π2t×12 ( sign not considered );A=8tπ
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