Homework 4

3-8

Here we use the fact that the spherical sound waves can be approximately expressed as $f(r,t) = {s\over r} \cos( kr-\omega t)$, and that $I={\cal P}/4\pi r^2 = {1\over 2}\rho v \omega ^2 s^2/r^2$. ${\cal P}_1 = 1.2 \times 10^{-3}~\rm W$ and ${\cal P}_2 = 1.8 \times 10^{-3}~\rm W$ (a) Assuming the sound velocity of $v= 343 ~\rm m/s$, the wavelength is $v/f = 1.72 ~\rm m$. The phase at $P$ from $S_1$ is $2\pi (4/1.72) =2\pi\times 2.32$, and from $S_2$, $2\pi (3/1.72)=2\pi \times 1.74$. Therefore, the phase difference is $\phi=0.58\times 2\pi$. (b) Since ${\cal P}_i = 2\pi \rho v \omega^2 s_i^2$ for $i=1,2$, from the power source $S_1$, we get $s_1 = 5.42\times 10^{-7} ~\rm m^2$. Likewise, $s_2 = 6.64\times 10^{-7} ~\rm m^2$. For $r_1 = 4~\rm m$ and $r_2 = 3~\rm m$, the wave function at $P$ is

$$f= {s_1\over r_1}\cos(\omega t +\phi) + {s_2\over r_2}\cos(\omega t)~.$$

The total intensity is

$$I ={1\over 2} \rho v \omega ^2 [ {s_1^2\over r_1^2}+ {s_2^2\over r_2^2} +2{s_1s_2\over r_1 r_2}\cos\phi ]$$

. (c) $I_2 = {\cal P}_2/4\pi r_2^2$. (d) $I_1 = {\cal P}_1/4\pi r_1^2$.

3-9

$L=15~\rm cm$, and at $n=1$, the wavelength is $\lambda = 2L = 30 ~\rm cm$. The speed on the wire is $v_w = 250 ~\rm m/s$ and that in the air is $v_s = 348 ~\rm m/s$. (a) The frequency $f = v_w/\lambda = 833~\rm Hz$. (b) $v_s/f = 42~\rm cm$.

3-10

(a) Since $f_n = nv/2L$, the fundamental frequency would be $f_1 = v/2L = 420-315= 105 ~\rm Hz$, which is the lowest resonant frequency. (b) $v = 2Lf_1 = 158 ~\rm m/s$.

3-11

Initially we do not know which fork is vibrating at a higher frequency although we do know that the frequency difference is $3~\rm Hz$. The fork of unknown frequency gains a small additional mass (wax), and we know that it decreases the natural frequency. This decreases the beating frequency, so we know that the unknown frequency is higher than that of the standard fork. $f=387~\rm Hz$.

3-12

$L=2.7~\rm m$, $M = 0.26~\rm kg$, and $T = 36~\rm N$. Then $v= \sqrt{T\over M/L}=190 ~\rm m/s$. The power transmission is ${\cal P}= {1\over 2}{M\over L}v\omega ^2 A^2= 85 ~{\rm W}$ and $A=7.7\times 10^{-3}~\rm m$. Then $f= 63~\rm Hz$.

3-13

$\beta = 10 \log (I/I_0)$ where $I_0 = 1.00 \times 10^{-12} ~\rm W/m^2$. (a) If $\beta = 70$, $I= 1.00 \times 10^{-5} ~\rm W/m^2$. (b) $I=\Delta P^2 /2\rho v$ so $\Delta P= \sqrt{2I\rho v}= 0.090 ~\rm Pa$.

3-14

If the sound source is moving, the observed frequency is $f^\prime = f v/(v\pm v_s)$. (a) The car must move away from her with a speed $v_s = v/2 = 170 ~m/s= 612 ~\rm km/h$. This is not a practical test at all, since the car need to run at about 5 times the normal speed. (b) $v_s = v/10 = 34 ~\rm m/s = 120 ~\rm km/h$. This is within the reach of a car speed.

3-15

The relative speed is $\Delta v = \Delta \lambda c/\lambda =1.94 \times 10 ^7 ~\rm m/s$, receding from the earth.

4-1

(a) From $T_F= {9\over 5}T_C + 32 ~\rm F$, $T=-40 \deg$. (b) From $T_F= {9\over 5}(T_K-273) + 32 ~\rm F$, $T = 573 \deg$. (c) Celsius and Kelvin never meet.

4-2

$\beta = {V_f-V_i \over V_i \Delta T}= 3\times 10^{-5} ~\rm K^{-1}$. For such a small expansion over $\Delta T= 2700 ~\rm K$, $\Delta V/V \approx 3\Delta R/R$. So $\Delta R = \Delta T \beta R/3 \approx 1.62 \times 10^5~\rm m$.