Question

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 it
oscillates through height of 1 m
Its maximum vertical velocity is 2 ms -1
Its maximum vertical velocity is 1 ms-1
Its maximum vertical acceleration is 16 ms
-2.
Its minimum vertical acceleration is 10 ms
-2​

Answer 1

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!♧