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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 uB. 6 uC. 12 uD. 3 u |
Answer» Correct Answer - B `(v_(m_(2)))/(v_(m_(1)))=(A_(2)omega_(2))/(A_(1)omega_(1))=(2T_(1))/(T_(2))=(2T_(1))/((1)/(3)T_(1))=3xx2` `v_(m_(2))=6v_(m_(1))` `=6xxu=6u` |
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