Answer
$(a)\space 2.63\times10^{5}N/m$
$(b)\space 2.09\space m/s$
$(c)\space 90\space m/s^{2}$
Work Step by Step
(a) We use the equation Frequency $(f) = \frac{1}{2\pi}\sqrt {\frac{K}{m}}$ to find the spring constant (K).
$f = \frac{1}{2\pi}\sqrt {\frac{K}{m}}$
$f^{2}=\frac{1}{4\pi^{2}}\times\frac{K}{m}$
$4\pi^{2}f^{2}m=K$; Let's plug known values into this equation.
$4\pi^{2}\times(6.85\space s^{-1})^{2}\times 142\space kg=K$
$263044.5\space N/m=K$
$2.63\times10^{5}N/m=K$
(b) We use the equation $V_{max}=A\omega$ to find the maximum acceleration.
$V_{max}=A\omega-(1)$
$\omega=2\pi f-(2)$
$(2)=\gt (1)$
$V_{max}=A\times2\pi f$ ; Let's plug known values into this equation.
$V_{max}=4.86\times10^{-2}m\times2\times\pi\times6.85\space s^{-1}$
$V_{max}=2.09\space m/s$
(c) We use the equation $a_{max}=\omega^{2}A$ to find the maximum acceleration.
$a_{max}=\omega^{2}A-(3)$
$(2)=\gt(3)$
$a_{max}=A\times (2\pi f)^{2}$ ; Let's plug known values into this equation.
$a_{max}=4.86\times10^{-2}\times4\pi^{2}\times6.85^{2}s^{-2}$
$a_{max}=90\space m/s^{2}$
