Problem 11.21 Challenge

กำหนดให้ \(T = 0.0200 s\) และ \(v = 20.0 m/s\)

จาก \(ω =\frac{2\pi}{T}= 100\pi\)

จาก \(v = \frac{\omega}{k}\)

\(k=\frac{\omega}{v}=5.00\pi\)

a) จาก \(y(x,t) = Asin(kx + ωt + φ) ;\) คลื่นเคลื่อนที่ไปทางทิศ -x

\(y(0,0) = Asin(k(0) + \omega (0) + φ)\)

\(0.0200 m = Asinφ\cdots \cdots (1)\)

จาก \(v_{\perp }=\frac{\partial y}{\partial t}\)

\(v_{\perp }=\frac{\partial}{\partial t}(Asin(kx + \omega t + φ))\)

\(v_{\perp }(x,t) = \omega Acos(kx + \omega t + φ)\)

\(v_{\perp }(0,0) = \omega Acos(k(0) + \omega (0) + φ)\)

\(-2.00 m = \omega Acos(φ)\cdots \cdots (2)\)

\((ω(1))^{2} + (2)^{2}\) จะได้

\((0.0200\omega )^{2} + (-2.00)^{2} = \omega ^{2}A^{2}\sin ^{2}φ + \omega ^{2}A^{2}\cos ^{2}φ\)

\((0.0200\omega )^{2} + (-2.00)^{2} = \omega ^{2}A^{2}(\sin ^{2}φ + \cos ^{2}φ)\)

\((0.0200\omega )^{2} + (-2.00)^{2} = \omega ^{2}A^{2}\)

\(A = \pm \frac{(0.0200\omega )^{2} + (-2.00)^{2}}{\omega}\)

\(A = 0.0210 m \;\blacksquare\)

\(b) (1)/(2)\) จะได้

\(\frac{0.0200}{-2.00}=\frac{Asinφ}{\omega Acosφ}\)

\(\tan φ = -0.01\omega\)

\(\tan φ = -\pi\)

\(φ = -1.26 rad \;\blacksquare\)

c) จาก \(v_{\perp }(x,t) = \omega Acos(kx + \omega t + φ)\)

จะได้ \(v_{\perp \max}= \omega A\)

\(v_{\perp \max}= 6.60 m/s \;\blacksquare\)

d) จาก \(y(x,t) = Asin(kx + ωt + φ)\)

จะได้ \(y(x,t) = (-2.10 cm)sin(5.00πx + 100πt - 1.26) \;\blacksquare\)