Answer
(a) T = 0.80 s
(b) f = 1.25 Hz
(c) $\omega = 7.85~rad/s$
(d) A = 3.0 cm
(e) $k = 148~N/m$
Work Step by Step
(a) The ball completes one cycle in 0.80 seconds. Therefore, the period $T$ is 0.80 seconds.
(b) We can find the frequency of the motion.
$f = \frac{1}{T} = \frac{1}{0.80~s} = 1.25~Hz$
(c) We can find the angular frequency.
$\omega = 2\pi~f$
$\omega = (2\pi)(1.25~Hz)$
$\omega = 7.85~rad/s$
(d) The ball oscillates back and forth between -3.0 cm and 3.0 cm. Therefore, the amplitude $A$ is 3.0 cm.
(e) We can find the force constant $k$
$\frac{k}{m} = \omega^2$
$k = \omega^2~m$
$k = (7.85~rad/s)^2(2.40~kg)$
$k = 148~N/m$
