Question

A particle is exceuting S.HM with a period of T seconds & Amplitude A m. The shortest time it takes to reach point A/√2 m​

Answer 1

SOLUTION.

If a body show SHM Than the position of the body after t time is given as...

X = A×Sin(wt)

where X is the position of the particle or body showing SHM

ACC to your question the at which position of particle is

X= A/√2

A/√2 = A×Sin(wt)

1/√2 = Sin(wt)

No. of solution = π/4, 3π/4

FIRST SOLUTION...

Sin(45) = Sin(wt)

wt = π/4

we know that w = 2π/T

(2π/T)×t = π/4

t= T/8

SECOND SOLUTION...

Sin(135) = Sin(wt)

wt = 3π/4

(2π/T)×t = 3π/4

t= 3T/8

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