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Solution :

Here, `a=20 "cm", omega = sqrt((k)/( m)) = sqrt((1200)/( 3)) = 20 "s"^(-1)` <br> a) As time is noted from the mean position, hence using `x=a sin omegat`, we have `x=2 sin 20t`. <br> b) At maximum stretched position, the body is the extreme right position, with an initial phase of `(pi)/(2)` rad. Then <br> `x=a sin (omega t + (pi)/(2))` <br> `= a cos omega t = 2 cos 20 t` <br> c) At maximum compressed position, the body is at the extreme left position, with an initial phase of `(3pi)/(2)` rad. <br> Then `x=a sin (omega t + (3pi)/(2))` <br> `= -a cos omega t` <br> ` = -2 cos 20 t`