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If the displacement of a particle executing S.H.M. is given by x = 0.24 sin (400 t + 0.5)m, then the maximum velocity of the particle isA. `24 m//s`B. `48 m//s`C. `96 m//s`D. `72 m//s` |
Answer» Correct Answer - C `x=0.24sin(400t+0.5)m` `v_(m)=Aomega` `=0.24xx400=96m//s` |
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