A particle executing simple harmonic motion has an angular frequency of 6.28 $s^{- 1}$ and an amplitude of 10 cm, the speed when the displacement is 6 cm from the mean position is:

50.2 c m / s

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60.2 c m / s

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70.2 c m / s

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80.2 c m / s

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Solution

The correct option is A $50.2 c m / s$
Velocity = angular frequency $\times \sqrt{(a m p l i t u d e)^{2} - (d i s p l a c e m e n t)^{2}} = 6.28 \times \sqrt{10^{2} - 6^{2}} = 6.28 \times 8 = 50.24 \frac{c m}{s}$

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A particle executing simple harmonic motion has an angular frequency of 6.28 s−1 and an amplitude of 10 cm, the speed when the displacement is 6 cm from the mean position is:

The correct option is A 50.2cm/sVelocity = angular frequency ×√(amplitude)2−(displacement)2=6.28×√102−62=6.28×8=50.24cms

A particle executing simple harmonic motion has an angular frequency of 6.28 $s^{- 1}$ and an amplitude of 10 cm, the speed when the displacement is 6 cm from the mean position is:

The correct option is A $50.2 c m / s$
Velocity = angular frequency $\times \sqrt{(a m p l i t u d e)^{2} - (d i s p l a c e m e n t)^{2}} = 6.28 \times \sqrt{10^{2} - 6^{2}} = 6.28 \times 8 = 50.24 \frac{c m}{s}$