Homework 3

2-11

Use the equation $P + \rho gh + {1\over 2}\rho v^2 = \rm constant$. (a) The pressure on the surface of the water (atmospheric pressure $P_0$) and at the hose are the same. So we compare the point $A$ and $C$,

$$P_0 + \rho g d - \rho gd =P_0 -\rho g (d+h_2) +{1\over 2} \rho v^2$$

. Then $v=\sqrt{2g(h_2+d)}$. (b) At point $B$, the velocity is $v$ obtained in (a), and the height is $h_1$. Compared to point $A$,

$$P_0 = P_B +\rho g h_1 + {1\over 2} \rho v^2$$

. So $P_B = P_0 -\rho g (h_1 +h_2 + d)$. (c) To lift water, the pressure needs to be greater than zero. $P_B >0$ leads to $h_1 < [P_0 - \rho g(h_2 +d)]/(\rho g)$. To get a maximum value, we set $h_2=d=0$ and use $P_0 = 1.015 \times 10^5 ~\rm Pa$. Then $h_1 < 10.3 ~\rm m$.

3-1

(a) and (b) omitted. (c) speed = 200 cm/s in the $-x$ direction.

3-2

Assume the cross-sectional are of the steel wire $A$. The maximum tension per area is $\tau = 7.0\times 10^8 ~\rm N/m^2$, so the maximum tension is $\tau A$. The linear mass density is $\rho A$. Therefore, the maximum speed is $v = \sqrt{ \tau /\rho} = 300 ~\rm m/s$.

3-3

(a) Zero. (b) 0.30 m.

3-4

Given $T=25 ~\rm ms$, $v_\phi =-30.0 ~\rm m/s$, $y(0,0) = 0.0200 ~\rm m$, and $v_y(0,0) = 2.00 ~\rm m/s$, we can obtain the angular frequency and the wave number: $\omega = 2\pi/T = 251 ~\rm s^{-1}$ and $k = \omega / v_\phi = 8.38 ~\rm m^{-1}$. We can first write down the wave function

$$y(x,t) = A\cos (kx+\omega t +\phi)$$

. (a) $y(0,0)=A\cos \phi$ and $v_y(0,0)=-\omega A\sin \phi$. So $A = \sqrt{ y^2(0,0)+ v_y^2(0,0)/\omega ^2} = 0.0215~\rm m$ (b) $y(0,0)/A =\cos \phi = 0.93$, and $-v_y(0,0)/(\omega A)= \sin \phi = 0.037$. Therefore, $\phi \approx 0.037$. (c) Maximum $v_y$ is $A\omega = 5.40 ~\rm m/s$. (d) Wave function: $y(x,t) = 0.0215 \cos (8.38 x + 251 t + 0.037)$.

3-5

$f =4.2 ~\rm GHz$, $v_\phi = 240 ~\rm m/s$. (a) The wavelength is $\lambda = 2\pi /k = v_\phi /f = 5.7\times 10^{-8} ~\rm m$. (b) $2~ \rm\mu m/\lambda = 35$ wavelengths. This is an enough resolution to take detailed images of the circuits.

3-6

$s(x,t) = A \cos( kx -\omega t)$. (a) amplitude: $A=2.00 ~\rm\mu m$. wavelength: $2\pi/k = 0.400 ~\rm m$. speed: $\omega/ k = 54.6 ~\rm m/s$. (b) $-1.15 ~\rm\mu m$. (c) Maximum speed: $A\omega = 1.72 ~\rm mm/s$.

3-7

The total wave can be expressed as

$$y(x,t) = y_m \cos(kx-\omega t )+y_m\cos(kx-\omega t +\pi/2) =2y_m \cos (\pi/4)\cos( kx-\omega t +\pi/4)$$

. So the total amplitude is $\sqrt{2} y_m$.