The maximum speed and acceleration of a particle executing simple harmonic motion are $10 c m s^{- 1} a n d 50 c m s^{- 2}$ . Find the position(s) of the particle when the speed is $8 c m^{- 1}$ .

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Solution

The correct option is B

Max speed

$v = A ω = 10$ ...(1)

max acceleration

$a = A ω^{2} = 50$ ...(ii)

Dividing (ii) with (i)

$ω = 5$ ...(iii)

substituting (iii) in (i)

$10 = A \times 5$

$A = 2$ ...(iv)
We derived the relation between v and x as.

$v = ω \sqrt{A^{2} - x^{2}}$

To find x when v = 8

$8 = 5 \sqrt{4 - x^{2}}$

$x^{2} = 4 - \frac{64}{25} x = \sqrt{\frac{36}{25}} = \pm \frac{6}{5} = \pm 1.2 c m$

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The maximum speed and acceleration of a particle executing simple harmonic motion are $10 c m s^{- 1} a n d 50 c m s^{- 2}$ . Find the position(s) of the particle when the speed is $8 c m^{- 1}$ .