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Question Number 133590 by physicstutes last updated on 23/Feb/21

A particle performs simple harmonic motion between two points

A and B which are 10 m apart on a horizontal straight line .

When the particle is 3 m away from the centre, O, of the line AB,

its speed is 8 {ms}^{- 1} . Find the least time required for the

particle to move from B to the midpoint of OA .

Answered by mr W last updated on 23/Feb/21

v = V_{0} \cos ω t w i t h V_{0} = s p e e d a t O

s = S_{1} \sin ω t w i t h S_{1} = O B = 5 m

T = p e r i o d = \frac{2 π}{ω}

a t t = t_{1} : s = 3 m, v = 8 m / s

8 = V_{0} \cos ω t_{1}

3 = 5 \sin ω t_{1}

\Rightarrow {(\frac{8}{V_{0}})}^{2} + {(\frac{3}{5})}^{2} = 1

\Rightarrow V_{0} = 10 m / s

o n t h e o t h e r s i d e : V_{0} = ω S_{1}

\Rightarrow ω = \frac{V_{0}}{S_{1}} = \frac{10}{8} = \frac{5}{4} 1 / s

\Rightarrow T = \frac{2 π}{ω} = \frac{8 π}{5} s

l e a s t t i m e f r o m B t o O = \frac{T}{4} = \frac{2 π}{5}

a t m i d p o i n t O A = C : s = \frac{O A}{2} = \frac{5}{2}

\frac{5}{2} = 5 \sin ω t_{2}

\Rightarrow ω t_{2} = \sin^{- 1} \frac{1}{2} = \frac{π}{6}

\Rightarrow t_{2} = \frac{4 π}{5 \times 6} = \frac{2 π}{15}

l e a s t t i m e f r o m B t o C :

\frac{T}{4} + t_{2} = \frac{2 π}{5} + \frac{2 π}{15} = \frac{8 π}{15} \approx 1.675 s

Commented by physicstutes last updated on 23/Feb/21

thank you sir