The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will beA. `2sqrt(3)` cmB. `sqrt(3)` cmC. 1 cmD. 2 cm

The amplitude of a executing `SHM` is `4cm` At the mean position the s

1.	The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will beA. `2sqrt(3)` cmB. `sqrt(3)` cmC. 1 cmD. 2 cm
Answer» Correct Answer - D (d) At mean position , velocity is maximum `v_("max") =omegaaimpliesomega=(v_("max"))/(a)=(16)/(4)=4` `therefore v=sqrt(a^(2)-y^(2))implies 8sqrt(3)=4sqrt(4^(2)-y^(2))` `192=16(16-y^(2))implies12=16-y^(2)impliesy=2 cm`

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The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will beA. `2sqrt(3)` cmB. `sqrt(3)` cmC. 1 cmD. 2 cm

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