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The maximum velocity of a particle executing S.H.M. is u. If the amplitude is doubled and the time period of oscillation decreases to (1/3) of its original value, then the maximum velocity will beA. 18 VB. 6 VC. 12 VD. 3 V |
Answer» Correct Answer - B `(v_(m_(2)))/(v_(m_(1)))=(A_(2)omega_(2))/(A_(1)omega_(1))=(A_(2))/(A_(1))xx(T_(1))/(T_(2))` `=2xx3=6` `v_(m_(2))=6v` |
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