Problem 8.20

a) พิจารณาที่ตำแหน่งสมดุลของวัตถุ

จาก \(ΣF_{y}= ma_{y}\)

\(kx - mg = 0\)

\(k =\frac{mg}{x}\)

จาก \(T = 2π\sqrt{\frac{m}{k}}\)

\(T = 2\pi \sqrt{\frac{mx}{mg}}\)

\(T = 2\pi \frac{x}{g}= 2\pi \sqrt{\frac{2.50 cm}{9.80 m/s^{2}}}= 0.317 s \;\blacksquare\)

b) จาก \(v_{\max}= \omega _{0}A =\frac{2\pi A}{T}=\frac{2\pi (5.00 cm)}{0.317 s}= 0.990 m/s \;\blacksquare\)

c) จาก \(a_{\max}= \omega _{0}^{2}A =\frac{4\pi ^{2}A}{T^{2}}=\frac{4\pi ^{2}(5.00 cm)}{(0.317 s )^{2}}= 19.6 m/s^{2} \;\blacksquare\)

d) จาก \(v = ω_{0}\sqrt{A^{2} - x^{2}}\)

\(v =\frac{2\pi}{T}\sqrt{A^{2} - x^{2}}\)

\(v =\frac{2\pi}{0.317 s}\sqrt{(5.00 cm)^{2} - (4.00 cm)^{2}}\)

\(v = 0.595 m/s \;\blacksquare\)

e) จาก \(a = -ω_{0}^{2}x\)

\(a = -(\frac{2\pi}{T})^{2}x\)

\(a = -(\frac{2\pi}{0.317 s})^{2}(4.00 cm)\)

\(a = -15.7 m/s^{2}\)

\(\therefore |a| = 15.7 m/s^{2} \;\blacksquare\)