The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is
0.5m $s^{- 1}$
1 m $s^{- 1}$
1.414m $s^{- 1}$
0.75m $s^{- 1}$

Question

The velocity of a particle that executes S-H-M- is its mean position 0-866 m-s-Velocity a displacement half of its amplitude from mean position is0-5m s-1- 1 m s-1- 1-414m s-1- 0-75m s-1-

Toppr Admin · Accepted Answer

-vmax-A-x3C9-0-866msAt x-A2-v-x3C9-x221A-A2-x2212-x2v-x221A-32-x3C9-A-x221A-32vmaxv-x221A-32-xD7-0-866-0-75ms

The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is0.5ms−11 ms−11.414ms−10.75ms−1

vmax=Aω=0.866msAt x=A2;v=ω√A2−x2v=√32ωA=√32vmaxv=√32×0.866=0.75ms

The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is
0.5m $s^{- 1}$
1 m $s^{- 1}$
1.414m $s^{- 1}$
0.75m $s^{- 1}$

$v_{m a x} = A ω = 0.866 \frac{m}{s}$
At $x = \frac{A}{2}; v = ω \sqrt{A^{2} - x^{2}}$
$v = \frac{\sqrt{3}}{2} ω A = \frac{\sqrt{3}}{2} v_{m a x}$
$v = \frac{\sqrt{3}}{2} \times 0.866 = 0.75 \frac{m}{s}$