Two particles of same period T and amplitude undergo SHM along the same line with initial phase of -6- If they start at the same instant and at same point along opposite directions- find the time after which they will meet again for the first time-

The correct option is A 3T/8 x=Asin ( wt π2 ) sin ( wt π2 )=sin wt 2wt π2=π 2wt=32π wt=34π t=34πw t=2πw ( 38 ) ∴ t= ...

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Byju's Answer

Standard XII

Physics

Time Period Independent of Amplitude

Two particles...

Question

Two particles of same period $(T)$ and amplitude undergo $S H M$ along the same line with initial phase of $π / 6$ . If they start at the same instant and at same point along opposite directions, find the time after which they will meet again for the first time:

3 T / 8

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T / 4

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Solution

The correct option is A $3 T / 8$
$x = A sin (w t - \frac{π}{2})$

$sin (w t - \frac{π}{2}) = sin w t$

$2 w t - \frac{π}{2} = π$

$2 w t = \frac{3}{2} π$

$w t = \frac{3}{4} π$

$t = \frac{3}{4} \frac{π}{w}$

$t = \frac{2 π}{w} (\frac{3}{8})$

$∴ t = \frac{3 T}{8}$

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Two particles of same period (T) and amplitude undergo SHM along the same line with initial phase of π/6. If they start at the same instant and at same point along opposite directions, find the time after which they will meet again for the first time:

The correct option is A 3T/8x=Asin(wt−π2)sin(wt−π2)=sinwt2wt−π2=π2wt=32πwt=34πt=34πwt=2πw(38)∴t=3T8

Two particles of same period $(T)$ and amplitude undergo $S H M$ along the same line with initial phase of $π / 6$ . If they start at the same instant and at same point along opposite directions, find the time after which they will meet again for the first time:

The correct option is A $3 T / 8$
$x = A sin (w t - \frac{π}{2})$

$sin (w t - \frac{π}{2}) = sin w t$

$2 w t - \frac{π}{2} = π$

$2 w t = \frac{3}{2} π$

$w t = \frac{3}{4} π$

$t = \frac{3}{4} \frac{π}{w}$

$t = \frac{2 π}{w} (\frac{3}{8})$

$∴ t = \frac{3 T}{8}$