A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceleration at a distance of 2 cm from the mean position is 8 cm/ $s^{2}$ . The maximum speed of the particle is

[EAMCET (Engg.) 2000]

8 cm/s

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12 cm/s

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16 cm/s

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24 cm/s

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Solution

The correct option is B
12 cm/s

A = $ω^{2} y \Rightarrow ω = \sqrt{\frac{A}{y}} = \sqrt{\frac{8}{2}} = 2 r a d / s e c$

Now $v_{m a x} = a ω = 6 \times 2 = 12 c m / s e c$

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A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceleration at a distance of 2 cm from the mean position is 8 cm/s2. The maximum speed of the particle is [EAMCET (Engg.) 2000]

The correct option is B 12 cm/s A = ω2y⇒ω=√Ay=√82=2rad/sec Now vmax=aω=6×2=12cm/sec

A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceleration at a distance of 2 cm from the mean position is 8 cm/ $s^{2}$ . The maximum speed of the particle is

[EAMCET (Engg.) 2000]

The correct option is B
12 cm/s

A = $ω^{2} y \Rightarrow ω = \sqrt{\frac{A}{y}} = \sqrt{\frac{8}{2}} = 2 r a d / s e c$

Now $v_{m a x} = a ω = 6 \times 2 = 12 c m / s e c$