The SHM is represented by `y= 3sin 314 t+ 4 cos 314t` y in cm and t in second. Find the amplitude of SHO.

Explanation

`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`.