Solutions

1

From the form of the wave-function, we see that the amplitude is $A= 0.1 ~\rm m$, the angular frequency is $\omega = 10 ~\rm rad/s$ and the phase constant is $\phi =\pi/2$. The mass is given $M= 0.1~\rm kg$.

(a)

$f=\omega / 2\pi = 1.6 ~\rm Hz$.

(b)

The maximum velocity is $v_M = \omega A = 1.0 ~\rm m/s$. This happens at $x=0$.

(c)

The maximum acceleration is $a_M = \omega^2 A = 10 ~\rm m/s^2$. This happens at the maximum displacement (in the $-x$ direction) $x=-0.1 ~\rm m$.

(d)

$F= Ma_M = 1.0~\rm N$.

(e)

The total energy can be obtained from the maximum kinetic energy $E= {1\over 2} Mv_M^2 = 0.05 ~\rm J$.

2

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)

Since the number of fluid particles is conserved, the equation of continuity holds, which says $Av =~\rm constant$. So $v$ is constant if the cross-sectional area $A$ is constant throughout the tube.

(c)

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 $P_B = P_0 -\rho g (h_1 +h_2 + d)$.

(d)

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 $h_2=d=0$. Then $h_1 < {P_0\over rho g}$.

(e)

You just have to add the acceleration $a$ to the gravity $g$. So $h_1< {P_0\over rho (g+a)}$.

3

(a)

$v= 30/2.0 = 15~\rm m/s$, in the $-x$ direction.

(b)

From $\sqrt{T/\mu} = v =15 ~\rm m/s$, you get $T= 3.6\times 10^{-2}~\rm N$.

(c)

If the total wave is of the form $y= A\cos(kx+\omega t)+B\sin (kx+\omega t)$, the time-averaged kinetic energy of a segment $\Delta x$ of the string is

$$\Delta\bar{K}= \int_0^T {dt\over T} {1\over 2}\mu \Delta x (dy/dt)^2 ={1\over 4}\mu \Delta x \omega^2 (A^2+B^2)~.$$

Then the total average power is ${\cal P} = 2 \Delta\bar{K}/\Delta t$. If you relate $\Delta x/\Delta t = v$, you get ${\cal P}={\mu\over 2} v\omega ^2 (A^2+B^2)= 3.2\times 10^{-3} ~\rm W$.

(d)

The fundamental mode is $f_1 = {v\over 2L}$. If you set up the inequality $20 {~\rm Hz} < f_1 < 20,000{~\rm Hz}$, you get $0.38 {~\rm m} > L > 3.8\times 10^{-4} ~\rm m$.

4

(a)

$f_1 \approx f_0(1-v_o/v) \approx 28 {~\rm kHz}-80 ~\rm Hz$.

(b)

$f_2 \approx f_1 (1-v_o/v) \approx 28 {~\rm kHz}-160 ~\rm Hz$.

(c)

The beat frequency is simply the frequency difference between $f_0$ and $f_2$, which is about $160 ~\rm Hz$.