A particle executes simple harmonic motion along a straight line with mean position at $x = 0$ and period of $20 s$ and amplitude of $5 c m$ . The shortest time taken by the particle to go from $x = 4$ cm to $x = - 3$ cm is

4 s

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7 s

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5 s

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6 s

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Solution

The correct option is C $7 s$
$x = A ($ sin $ω t + ϕ$ )
$ω = \frac{2 π}{20} = \frac{π}{10} \frac{r a d}{s e c}$
let at t=0, x=4, thus
$4 = 5$ sin $ϕ$
$s i n^{- 1} \frac{4}{5} = ϕ$
Now for at $t = t_{1}$ , let x=-3, thus we have
$\frac{- 3}{5} = s i n (ω t_{1} + ϕ)$
or
= $s i n^{- 1} (\frac{- 3}{5}) - s i n^{- 1} (\frac{4}{5}) = ω t_{1}$

= $10 \frac{(- 0.643 - 0.927)}{π} = t_{1}$
Solving we get $t_{1} = 5 s e c$

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A particle executes simple harmonic motion along a straight line with mean position at x=0 and period of 20 s and amplitude of 5 cm. The shortest time taken by the particle to go from x=4cm to x=−3cm is

The correct option is C 7 sx=A(sin ωt+ϕ)ω=2π20=π10radseclet at t=0, x=4, thus4=5 sin ϕsin−145=ϕNow for at t=t1, let x=-3, thus we have−35=sin(ωt1+ϕ)or=sin−1(−35)−sin−1(45)=ωt1=10(−0.643−0.927)π=t1Solving we get t1=5sec

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