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The vertical motion of a ship at sea is described by the equation `(d^(x))/(dt^(2))=-4x` , where x is the vertical height of the ship (in meter) above its mean position. If it oscillates through a height of 1 mA. its maximum vertical speed will be `1 ms^(-1)`B. its maximum vertical speed will be `2 ms^(-1)`C. its greater vertical acceleration is `2 ms^(-2)`D. its greater vertical acceleration is `1 ms^(-2)` |
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Answer» Correct Answer - B (b) Comparing with `(d^(2)x)/(dt^(2))=-omega^(2)x` we have , `omega=2 "rad"s^(-1)` `v_("max")=omegaA=2ms^(-1)` `a_("max")=omega^(2)A=4 ms^(-2)` |
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