A particle undergoes SHM according to X - A sin wt - - 3- What would be the shortest time taken by it to move from X - A - 2 to X - -A - 2 -A- -2 wB- -6 wC- -3 wD- -4 w

The correct option is C π/3wUsing phase circle, t= θ/w = π /6+ π /6/w = π /3/w = π/3w ...

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Standard XII

Physics

Horizontal and Vertical Components of Projectile Motion

A particle un...

Question

A particle undergoes SHM according to $X = A s i n (w t + π / 3$ ). What would be the shortest time taken by it to move from $X = A / 2$ to $X = - A / 2$ ?

\frac{π}{2 w}

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\frac{π}{3 w}

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\frac{π}{4 w}

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\frac{π}{6 w}

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Solution

The correct option is B $\frac{π}{3 w}$
Using phase circle,
$t = \frac{θ}{w} = \frac{π / 6 + π / 6}{w} = \frac{π / 3}{w} = \frac{π}{3 w}$

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A particle undergoes SHM according to X=A sin(wt+π/3). What would be the shortest time taken by it to move from X=A/2 to X=−A/2 ?

The correct option is B π3wUsing phase circle, t=θw=π/6+π/6w=π/3w=π3w

A particle undergoes SHM according to $X = A s i n (w t + π / 3$ ). What would be the shortest time taken by it to move from $X = A / 2$ to $X = - A / 2$ ?

The correct option is B $\frac{π}{3 w}$
Using phase circle,
$t = \frac{θ}{w} = \frac{π / 6 + π / 6}{w} = \frac{π / 3}{w} = \frac{π}{3 w}$