Chapter 21 - Superposition - Exercises and Problems - Page 626: 79

See the detailed answer below.

$$\color{blue}{\bf [a]}$$ Since these waves resemble a standing sound wave in an open-open tube, the wavelength is then given by $$\lambda_m=\dfrac{2L}{m}\tag{$m=1,2,3,...$}$$ So the first three waves are, At $m=1$, $$\lambda_1=\dfrac{2(10)}{1}=\color{red}{\bf 20}\;\rm m$$ The water depth here is just one-quarter the wavelength. So we can barely consider it as a deep-water wave. At $m=2$, $$\lambda_2=\dfrac{2(10)}{2}=\color{red}{\bf 10}\;\rm m$$ The water depth here is greater than one-quarter the wavelength. So it is a deep-water wave. At $m=3$, $$\lambda_3=\dfrac{2(10)}{3}=\color{red}{\bf 6.67}\;\rm m$$ The water depth here is greater than one-quarter the wavelength. So it is a deep-water wave. See the figures below. $$\color{blue}{\bf [b]}$$ The wave speed is given by $$v=\sqrt{\dfrac{g\lambda}{2\pi}}$$ So, $$v_1=\sqrt{\dfrac{(9.8)(20)}{2\pi}}=\color{red}{\bf 5.59}\;\rm m/s$$ $$v_2=\sqrt{\dfrac{(9.8)(10)}{2\pi}}=\color{red}{\bf 3.95}\;\rm m/s$$ $$v_3=\sqrt{\dfrac{(9.8)(\frac{20}{3})}{2\pi}}=\color{red}{\bf 3.22}\;\rm m/s$$ $$\color{blue}{\bf [c]}$$ We know that the frequency is given by $$f=\dfrac{v}{\lambda}$$ So, $$f_m=\dfrac{\sqrt{\dfrac{g\lambda_m}{2\pi}}}{\lambda_m}=\sqrt{\dfrac{g}{2\pi \lambda_m}}$$ where $\lambda_m=2L/m$ $$\boxed{f_m =\sqrt{\dfrac{gm}{4\pi L}}}$$ $$\color{blue}{\bf [d]}$$ We know that the period is given by $$T=\dfrac{1}{f}$$ where $f=\sqrt{\dfrac{gm}{4\pi L}}$ So, $$T=\dfrac{1}{\sqrt{\dfrac{gm}{4\pi L}}}$$ Hence, $$T=\dfrac{1}{\sqrt{\dfrac{m(9.8)}{4\pi (10)}}}$$ Therefore, $$T_1=\dfrac{1}{\sqrt{\dfrac{(1)(9.8)}{4\pi (10)}}}=\color{red}{\bf 3.58}\;\rm s$$ $$T_2=\dfrac{1}{\sqrt{\dfrac{(2)(9.8)}{4\pi (10)}}}=\color{red}{\bf 2.53}\;\rm s$$ $$T_3=\dfrac{1}{\sqrt{\dfrac{(3)(9.8)}{4\pi (10)}}}=\color{red}{\bf 2.07}\;\rm s$$