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The amplitude of a damped oscillator becomes half in one minute. The amplitude after three minutes will be (1)/(x) times the original, then find the value of x. |
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Answer» Solution :Amplitude of damped OSCILLATION at time t, `A(t) = A_(0)e^(-(bt)/(2m))` After `t= 1 MIN, A(2) = (A_0)/(2)` `therefore (A_0)/(2) = A_(0)e^(-(b(1))/(2m))` `therefore (1)/(2)= e^(-(b)/(2m))` `therefore 2 = e^(-(b)/(2m))"""........."(1)` Now, after `t= 3 min, A(3) = (A_0)/(x)` `therefore (A_0)/(x) = A_(0)e^(-(b(3))/(2m))` `therefore (1)/(x)= e^(-(3b)/(2m))` `therefore x = e^((3b)/(2m))= (e^((b)/(2m)))^(3)` `therefore x= 2^(3)` (from equation (1)). |
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