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The SHM is represented by y= 3sin 314 t+ 4 cos 314t y in cm and t in second. Find the amplitude of SHO. |
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Answer» Solution :`y= 3sin 314t + 4 cos 314t` shows SHM. We can have displacement of SHM at time t from the quation `y= A SIN (OMEGA t+phi)`. `y= A sin omega t cos phi + A cos omega t sin phi` In `y= (A cos phi) sin omega t+(A sin phi) cos omega t" putting "omega = 314 rad s^(-1)`. COMPARING equation `y= (A cos phi) sin 314 t+ (A sin phi) cos 314t` with given equation `y= 3sin 314 t + 4 cos 314t`. `3= A cos phi " and "4= A sin phi"""........"(1)` `therefore (3)^(2) +(4)^(2) = A^(2) cos^(2) +A^(2) sin^(2) phi` `therefore 9+ 16 =A^(2) (cos^(2) phi+ sin^(2) phi)` `therefore A^(2)= 25 implies A = pm 5 cm`. Now from equation (1), `(4)/(3) = tan phi` `therefore phi = tan^(-1) (1.3333)` `phi = 53.8^(@)` and periodic time `T= (2pi)/(omega)` `=(2xx 3.14)/(314) implies T= 0.02 s` and maximum VELOCITY, `v_("max") = A omega = 5xx 314` `therefore v_("max") = 1570 cm"/"s`. |
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