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The average velocity of a particle executing SHM with an amplitude A and angular frequency `omega` during one oscillation isA. `omegaA`B. `(omegaA)/(2)`C. `2omegaA//pi`D. zero |
Answer» Correct Answer - D `v_(1)=omegasqrt(A^(2)-x_(1)^(2))` `3=omegasqrt(25-16)` `3=3omega" "therefore omega=1` `T=(2pi)/(omega)=(2pi)/(1)=6.28s` |
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