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Asked by asianpaints808 | 04 Oct, 2020, 12:23: PM
Let t1 be the time taken by the bead to travel in slated side of wire frame.
Let t2 be the time taken by the bead to travel the horizontal side of frame.

Total time T1 for the bead to complete one cycle ( starting point to starting point ) is given as

T1 = 4t1 + 2t2    ................................ (1)

time to travel the vertical part is determined from the relation , h = (1/2)gt2 .

Hence if vertical part of wire frame becomes 4 times , then time is doubled.

Hence now the time taken by the bead to travel in slated side of wire frame is ( 2 t1 )

speed of bead when it reaches the horizontal part is determined from the relation , v2 = 2gh

Hence if vertical part of wire frame becomes 4 times , then speed in horizontal part is doubled.

Hence now the time taken by the bead to travel in horizontal part of wire frame is ( 2 t2 ) .

Total time T2 for the bead to complete one cycle after changing the linear dimension 4 times  is given as

T2 = 8t1 + 4t2 = 2 T1

Since Time period is doubled, frequency will become half.
Answered by Thiyagarajan K | 14 Oct, 2020, 12:45: AM

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