Answer
(a) $m = 0.25~kg$
(b) $A = 1.2~cm$
(c) $F_{max} = 3.0~N$
Work Step by Step
(a) From the graph, we can see that one cycle is completed in 0.20 seconds. Therefore, the period $T = 0.20~s$. We can find the mass $m$.
$T = 2\pi\sqrt{\frac{m}{k}}$
$m = \frac{T^2~k}{(2\pi)^2}$
$m = \frac{(0.20~s)^2(250~N/m)}{(2\pi)^2}$
$m = 0.25~kg$
(b) From the graph, we can see that $a_{max} = 12.0~m/s^2$. We can find the amplitude $A$.
$a_{max} = \omega^2~A = \frac{k~A}{m}$
$A = \frac{m~a_{max}}{k}$
$A = \frac{(0.25~kg)(12.0~m/s^2)}{250~N/m}$
$A = 0.012~m = 1.2~cm$
(c) We can find the maximum force the spring exerts on the glider.
$F_{max} = m~a_{max}$
$F_{max} = (0.25~kg)(12.0~m/s^2)$
$F_{max} = 3.0~N$
