The vertical motion of a ship at sea is described by equation of --4y,

1.	The vertical motion of a ship at sea is described by equation of --4y,where y is vertical height of ship in metre above its mean position. If itoscillates through height of 1 mIts maximum vertical velocity is 2 ms -1Its maximum vertical velocity is 1 ms-1Its maximum vertical acceleration is 16 ms-2.Its minimum vertical acceleration is 10 ms-2
Answer» Answer: ☆Concept: 》in the SHM, the object is doing to-and-fro motion due to restoring force directed towards the mean position. 》The maximum acceleration of the particle will be at extremes. And acceleration will be zero at mean position. $\boxed{\bf{ a= - \omega^{2}y}}$ 》maximum velocity will at mean position. $\boxed{\bf{V_{max} = \omega A}}$ ..(Ais the amplitude) ☆solution: 》 $\dfrac{d^{2}y}{dt^{2}}$ = a (double DIFFERENTIATION of displacement) 》a = -4y from above formula, $\omega^{2}$ = 4 $\omega$ = 2rad/s ---------- 》so, here the RANGE where it ossilates is given as 1m, that means its the DISTANCE between the two extremes = 2A 》A=1/2 = 0.5 --------- $\rightarrow{ V_{max} = \omega A}$ $\rightarrow{ V_{max} = 4×0.5}$ $\implies{\bf{ V_{max} = 2m/s}}$ ▪︎▪︎▪︎ $\rightarrow{ a_{max} = -\omega^{2}A}$ $\rightarrow{ a_{max} = -4×4×0.5}$ $\implies{\bf{ a_{max} = -8ms^{-2}}}$ (-ve shows the OPPOSITE direction) so ,from above 1st option is correct. ♧hope it helps!♧