Answer
a) $t=1.0s$, $3.0s$, $5.0s$
b) $F_{max}=4.68N$
c) $t=0$, $t=2.0$ ,$t=4.0$, $t=6.0s$.
d) $F=3.31N$
Work Step by Step
(a) We know that the maximum displacement occurs at times $t=1.0s$, $3.0s$, $5.0s$. Thus, when the force is applied at its maximum, the displacement will also be at a maximum.
(b) We know that
$F_{max}=mA(\frac{2\pi}{T})^2$
We plug in the known values to obtain:
$F_{max}=(3.8Kg)(0.50m)(\frac{2\times 3.14}{4.0s})^2$
$F_{max}=4.68N$
(c) We know that when the displacement is zero then the force applied will also be zero; the displacement is zero at times $t=0$, $t=2.0$ ,$t=4.0$, $t=6.0s$.
(d) We know that
$x=Asin(\omega t)$
$\implies x=Asin(\frac{2\times 3.14\times 0.5s}{4.0s})$
$\implies x=\frac{A}{\sqrt 2}=0.707A$
We know also know that
$F_{max}=ma_{max}$
$\implies F_{max}=m(A\omega^2)$
$\implies F_{max}=KA$
Now the required force exerted on the mass can be calculated as
$F=Kx$
$\implies F=0.707KA$
$\implies F=0.707F_{max}$
$\implies F=(0.707)(4.68N)$
$F=3.31N$
