Solved Problems

(a) Since both ends of the string are fixed, the nodes are formed at these points. Between any two consecutive nodes, there is half a wavelength. Therefore, in general, between any two nodes, we obtain $n$ half a wavelength that is $L=n\ \lambda/2$
Where $n=1,2,3,\dots$ represents the nth harmonic mode or the number of nodes. So the wavelength of nth mode and its frequency are given by the following relations 
$$\ \lambda_n=\frac{2L}{n}\ ,\ \ f_n=\frac{v}{\lambda_n}=n\frac{v}{2L}$$ 
(b) $$f_n=\frac{nv}{2L}=\frac{n\left(344\right)}{2\left(2.5\right)}=n\left(68.8\right)=\left(68.8,\ 137.6,206.4\right)\ {\rm Hz}$$
(c) In this case, one end of the pipe is open and the other is close (fixed) so at the open-end forms an anti-node and the fixed end forms a node. Between any consecutive node and anti-node, there is one fourth of a wavelength that is in general 
$$L=n\frac{\lambda}{4}\Rightarrow \lambda_n=\frac{4L}{n}\ \ and\ f_n=\frac{v}{\lambda_n}=n\frac{v}{4L}$$ 
But here, $n=1,3,5,\dots$ (the number of nodes). The three consecutive modes are

$$f_n=n\left(\frac{344}{4\times 2.5}\right)=n\left(34.4\right)\Longrightarrow \left(34.4\ ,\ 103.2\ ,\ 172\right)\ {\rm Hz}$$ 
Note: the audible frequencies are between $20\, {\rm Hz}$ and $20000\, {\rm Hz}$. Therefore, in both cases, the frequencies are audible. 

The speed of waves on a string with tension $T$ and linear mass density $\mu$ is given by 
$$v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{T}{\frac{m}{l}}}$$ 
We can suppose that all of mass of the rope is concentrated in the middle of it, so its tension is $T=\frac{mg}{2}$
$$v=\sqrt{\frac{\frac{mg}{2}}{\frac{m}{L}}}=\sqrt{\frac{gl}{2}}=4.7\ {\rm m/s}$$ 
 

(a) The general form of a wave equation is $D\left(x,t\right)=A{\cos  (kx-\omega t+\phi)\ }$ Where $A$ is the amplitude of the wave. Using initial conditions, we find the constants of the above equation. 
$$D\left(0,0\right)=5\to 5\,{\cos  \phi\ }=5\to \ \phi=0$$ 
$$k=\frac{2\pi}{\lambda}=\frac{2\pi}{2}=\pi$$ 
$$\omega=\frac{2\pi}{T}=\frac{2\pi}{\frac{v}{\lambda}}=50\pi$$ 
$$\therefore D\left(x,t\right)=0.005 \cos (\pi x-50\pi t)\$$ 

(b) The acceleration of a travelling wave is defined by the following relation
$$a=\frac{d^2}{dt^2}D\left(x,t\right)=-0.005\ {\left(50\pi\right)}^2{\cos  \left(\pi x-50\pi t\right)\ }$$ 
The above equation is maximum when cosine term be one. Then $a_{max}=120\ {\rm m/}{{\rm s}}^{{\rm 2}}$.

Note: wavelength defined as the distance between two successive picks or valleys. In this figure, there are three valleys or anti-nodes. Each valley is equal to half a wavelength so $$L=3\frac{\lambda}{2}\Rightarrow \ \lambda=\frac{2}{3}L$$ Use the relation between wavelength and frequency to find the speed of the wave on the string $$v=\lambda f=\frac{2}{3}Lf=\frac{2}{3}\left(0.6\right)\left(690\right)=276\frac{{\rm m}}{{\rm s}}$$ Another relation for speed of the waves on the string is $=\sqrt{\frac{T_S}{\mu}}$ , where $T_s$ is the tension on the string and $\mu=M/L$ is the linear mass density of it. So
$$\begin{align*}v=\sqrt{\frac{T_S}{\mu}}\Rightarrow T_S&=\mu v^2\\ &=\frac{M}{L}v^2\\ &=\frac{0.001}{0.6}{\left(276\right)}^2\\ &=126.96\ {\rm N}\end{align*}$$

(a) First, find the difference path of two speakers then use the relation $\Delta D=n\lambda\ ,\ n=1,2,\dots$ for instructive and $\Delta D=n\frac{\lambda}{2}\ ,\ n=1,3,\dots$ for destructive interference.
Let $x$ be the position of microphone along the $BC$ path.
$$\Delta D=\sqrt{x^2+2^2}-x=n\frac{\lambda}{2}$$ 
$$\begin{align*}x^2+4&={\left(n\frac{\lambda}{2}+x\right)}^2\\ &={\left(\frac{n\lambda}{2}\right)}^2+x^2+n\lambda x \end{align*}$$
$$\begin{gather*} \to n\lambda x=4-{\left(n\frac{\lambda}{2}\right)}^2\\ \Longrightarrow x\left(n\right)=\frac{4}{n\lambda}-n\frac{\lambda}{4}\ ,\ n=1,3,5,\dots \end{gather*}$$
$$\lambda=\frac{v}{f}=\frac{344}{784}=0.439\ {\rm m}$$ 
Let $n=1\Rightarrow x=\frac{4}{\left(1\right)\left(0.439\right)}-\frac{1\left(0.439\right)}{4}=9\ {\rm m}$
$n=3\to x=2.7\ {\rm m}$ and so on.
(b) Same procedure as part (a) : $\Delta D=n\lambda\ ,\ n=1,2,\dots \ \ \ \Rightarrow \ x\left(n\right)=\frac{4}{n\lambda}-n\frac{\lambda}{4}\$
(c) Because in this case there is no position along $BC$ so
$$x=0=\frac{4}{n\lambda}-n\frac{\lambda}{4}\to \ \lambda^2=\frac{16}{n^2}$$ 
$$n=1\ \Longrightarrow \ \lambda^2_{max}=16\to \lambda_{max}=4\ {\rm m}$$ 
$$f=\frac{v}{\lambda}=\frac{344}{4}=86\ {\rm Hz}$$ 

(a) You are asked about $y_{min}$, so we need destructive interference, recall that for occurring such interference, the path lengths taken by the two light rays has to differ by a half-integer number of wavelengths.

$$\Delta L=d\,{\sin  \theta_{min}\ }=\left(m+\frac{1}{2}\right)\lambda\ ,\ m=0,1,2,\dots \ \ \ and\ \ y=D\,{\tan  \theta\ }$$ 
Where $y$ , $d$ and $\theta$ are shown in the figure.
Solving for ${\sin  \theta_{min}\ }$, we obtain
$${\sin  \theta_{min}\ }=\left(m+\frac{1}{2}\right)\frac{633\times {10}^{-9}}{0.07\times {10}^{-9}}$$
$$\to \left\{$$$$\begin{array}{lcr} \theta_{min}\left(m=0\right)=0.259{}^\circ \ \ &,& \ \ y_{min}=13.6\, {\rm mm} \\ \theta_{min}\left(m=1\right)=0.777{}^\circ \ \ &,& \ \ y_{min}=40.7\, {\rm mm} \\ \theta_{min}\left(m=2\right)=1.30{}^\circ \ \ &,& \ \ y_{min}=68.1\, {\rm mm} \end{array}$$\right. Due to the completely destructive pattern, intensity at these positions is zero that is $I_{min\ }=0$
(b) Similar above: maxima points are constructive interference $\Delta L=d\,{\sin  \theta_{max}\ }=m\lambda\ ,\ m=0,1,2,\dots$
$${\sin  \theta_{max}\ }=\frac{m\lambda}{d}\to \left\{$$$$\begin{array}{lcr} \theta_{max}\left(m=0\right)=0{}^\circ \ \ &,&\ \ y_{max}=0 \\ \theta_{max}\left(m=1\right)=0.518{}^\circ \ \ &,&\ \ y_{max}=27.1\, {\rm mm} \\ \theta_{max}\left(m=2\right)=1.04{}^\circ \ \ &,&\ \ y_{max}=54.5\, {\rm mm} \end{array}$$ \right. 
Note: the intensity $I$ of a sinusoidal wave with amplitude $E_0$ is given by $I=\frac{1}{2}c\,\epsilon_0E^2_0$
In the interference pattern (two slit) at each point on the screen, there are two waves. In the case of instructive pattern, these waves are in phase and so 
$$\begin{align*}I_{max}&=\frac{1}{2}c\,\epsilon_0{\left(E_0+E_0\right)}^2\\ &=4\frac{1}{2}\epsilon_0\,cE^2_0\\ &=4I_0=4\left(200\frac{{\rm W}}{{{\rm cm}}^{{\rm 2}}}\right)\\&=800\ {\rm W/}{{\rm cm}}^{{\rm 2}}\end{align*}$$
In general, the intensity pattern for double slit experiment is $I=4I_0\,{{\cos }^2 \left(\frac{\pi d}{\lambda}\,{\sin  \theta\ }\right)\ }$ so 
$$\begin{gather*}I_{\frac{1}{4}}=\frac{1}{4}I_{max}\\ \Rightarrow 4I_0\,{{\cos }^2 \left(\frac{\pi d}{\lambda}{\sin  \theta_{\frac{1}{4}}\ }\right)\ }=\frac{1}{4}\left(4I_0\right)\\ \to 4\,{{\cos }^2 \left(\frac{\pi d}{\lambda}\,{\sin  \theta_{\frac{1}{4}}\ }\right)\ }=1\end{gather*}$$
$$\Rightarrow \theta_{\frac{1}{4}}={\arcsin  \left(\frac{\lambda}{\pi d}\ {\arccos  \left(\frac{1}{\sqrt{4}}\right)\ }\right)\ }=3.01\times {10}^{-3}\ {\rm rad}$$ 
Substituting it into the relation $y=D\,{\tan  \theta\ }$, we get
$$y_{\frac{1}{4}}=D\,{\tan  \theta_{\frac{1}{4}}\ }=9.04\ {\rm mm}$$ 

(a) Note: when an unpolarized light passes through a polarizer, the intensity of the transmitted light is exactly half that of the incident wave. So $I_A=\frac{I_0}{2}=\frac{15}{2}=7.5\ {\rm W/}{{\rm m}}^{{\rm 2}}$

(b) When polarized light passes through a polarizer (analyzer), the intensity of the transmitted light is $I=I_0{{\cos }^2\phi\ }$, where $\phi$ is the angle between the polarization direction of the incident light and the polarizing axis of the polarizer (Malus's law). So 
$$\begin{align*} I_B&=I_A\,{{\cos }^2 \phi} \\ 3&=7.5\,{{\cos}^2 \phi}\\ \Rightarrow {{\cos}^2 \phi}&=0.4\\ \Rightarrow \ {\cos\phi}&=\pm 0.633 \end{align*}$$
Therefore, we the angle is $$\phi=50.8{}^\circ$$ 

(a) Maximum and minimum are instructive and destructive points, respectively.
$d\,{\sin  \theta\ }=m\lambda\$for instructive and $d\,{\sin \theta\ }=\left(m+\frac{1}{2}\right)\lambda$ for destructive where $m=0,1,2,\dots$
$$\left.$$$$\begin{array}{lccr} {\rm For\ }\lambda_1 : & d\,\sin \theta=m_1\lambda_1\ &, \ max \\ {\rm For\ }\lambda_2\: & d\,{\sin  \theta\ }=\left(m_2+\frac{1}{2}\right)\lambda_2\ &,\ min \end{array}$$ \right\}\Rightarrow m_1\lambda_1=\left(m_2+\frac{1}{2}\right)\lambda_2
$$\to \frac{\lambda_1}{\lambda_2}=\frac{m_2+\frac{1}{2}}{m_1}$$  where $m_1 , m_2$ are integers.   
$$\begin{gather*} \frac{450}{600}=\frac{\lambda_1}{\lambda_2}=\frac{m_2+\frac{1}{2}}{m_1}\\ \\ \Rightarrow \frac{3}{4}=\frac{2m_2+1}{2m_1}\to \left\{ \begin{array}{c}2m_1=4\to m_1=2 \\ 2m_2+1=3\to m_2=1 \end{array}\right.\end{gather*}$$ Bright fringes are located at $y_m=L\,{\tan  \theta_m\ }$ or $y_m=L\frac{m\lambda}{d}$.$$y_m=L\frac{m\lambda}{d}=\left\{$$$$\begin{array}{lcr}L\displaystyle{\frac{m_1 \lambda_1}{d}}=4\frac{2\times 450}{0.2\times {10}^{-3}}=0.018\,{\rm m=1.8\ cm} \\ \\ \displaystyle{L\frac{\left(m_2+\frac{1}{2}\right)\lambda_2}{d}}=4\frac{\left(1+\frac{1}{2}\right)\times 600}{0.2\times {10}^{-3}}=0.018\ {\rm m}=1.8\ {\rm cm} \end{array}$$\right. 
As expected!

(b) $\left. \begin{array}{c} m_1=2\to 3^{nd}\ bright\ fringe \\ m_2=1\ \to 2^{nd}\ dark\ fringe \end{array}\right\}$  

(a) Recall that bright fringes are given by equation $d\,{\sin \theta}=m\lambda\ ,\ m=1,2,\dots$. We know from mathematics that the magnitude of ${\sin \theta}$ is less than or equal one i.e. $\left|{\sin  \theta\ }\right|\le 1$. Combine these notes to find the maximum value of the number of bright fringes $m$ $$\begin{gather*} {\sin \theta}=\frac{m\lambda}{d}\le 1 \\ \\ \Rightarrow \quad m\le \frac{d}{\lambda}=\frac{0.0116\times {10}^{-3}}{585\times {10}^{-9}}=19.8 \\ \\ \Rightarrow \quad m_{largest}=19\end{gather*}$$ 
By substituting the largest $m$ in the interference equation, we obtain the largest value of ${\sin  \theta\ }$.
$$\begin{align*} (\sin \theta)_{max}&=\frac{m_{max}\lambda}{d}\\\\&=19\,\frac{(5.85\times {10}^{-7})}{0.0116\times {10}^{-3}}\\ \\&=\pm 0.958\end{align*}$$
(b) $m_{total}=19$ indicates that on each side of the central fringe there is $19$ bright fringes so $$2\left(19\right)+1\left(central\ fringe\right)=39$$ 
(c) $$\begin{align*} {\sin \theta_{19}}&=\frac{m\lambda}{d}\\\\&=19\left(\frac{5.85\times {10}^{-7}}{0.0116\times {10}^{-3}}\right)\\\\&=\pm 0.958 \to \theta\\\\&=73.3{}^\circ \quad \text{most distant fringe}\end{align*}$$ And $$\begin{align*} {\sin \theta_1}&=\frac{m^{'}\lambda}{d}\\ \\&=1\left(\frac{5.85\times {10}^{-7}}{0.0116\times {10}^{-3}}\right)\\\\ &=2.89{}^\circ \quad \text{most near fringe}\end{align*}$$ 

Note: if an unpolarized light with intensity $S_0$ incident on a polarizer then the intensity of the transmitted light is $\frac{S_0}{2}$. This light is polarized so if a second polarizer is placed in front of it then the intensity becomes $S=S_0\,{{\cos }^2 \phi\ }$, where $\phi$ is the angle between polarizing axes of first and second polarizers. 
(a) As noted above: $S_A=\frac{S_0}{2}$ and $S_B=S_A\,{{\cos }^2 60{}^\circ \ }=\left(\frac{S_0}{2}\right){\left(\frac{1}{2}\right)}^2=\frac{S_0}{8}$
(b) First $P_3$ is placed in A so the average intensity of light emerges to the right of it is 
$$S_A=\frac{S_0}{2}\,{{\cos }^2 \phi^{'}\ }$$ 
Because we want no intensity after placing $P_3$ therefore $S_A$ must be zero.
$$\frac{S_0}{2}\,{{\cos }^2 \phi^{'}\ }=0\Rightarrow \ \phi^{'}=90{}^\circ$$ 
Now $P_3$ is placed into B. light passes through two polarizer before incident on it so 
$$\begin{gather*} S_B=\frac{S_0}{8}\,{{\cos }^2 \phi''}=0 \\ \left(no\ light\ transmitted\right) \Rightarrow \ \phi''=90{}^\circ \end{gather*}$$
Where $\phi^{''}$ is the angle between polarizing axes of $P_3$ and $P_2$. In the other hand, the angle between $P_1$ and $P_2$ is $60{}^\circ$. So the angle between $P_1$ and $P_3$ is $30{}^\circ$.  

The two cords produce a pulling upward force of $-kx$, so treat them as one spring with an effective spring constant $k$.
At equilibrium, the force of the cords is equal to the force of gravity $$\Sigma F_y=0\Rightarrow mg=kx\Rightarrow k=\frac{mg}{x}\ \ ,\ \ (*)$$ 
We know that the frequency of a body with mass $m$ that undergoes a simple harmonic motion is $f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\ \sqrt{\frac{k}{m}}$. So using (*)
$$\begin{align*} f=\frac{1}{2\pi}\ \sqrt{\frac{k}{m}}&=\frac{1}{2\pi}\ \sqrt{\frac{\frac{mg}{x}}{m}}\\\\ &=\frac{1}{2\pi}\sqrt{\frac{g}{x}}\\\\ &=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.05}}\\\\&=2.2\quad {\rm Hz}\end{align*}$$

(a) First determine the spring constant:
Without the pebble, the block and spring is in a mechanical equilibrium (motionless in $y$ direction) so
$$\Sigma F_y=0 \ \to \ m_Bg=kx_1$$ But the displacement of the spring ($x$) in this case is not given. When the pebble is placed on the block, instantaneously a new equilibrium position created, so $k\left(x_1+x_2\right)=\left(m_B+m_p\right)g$, where $x_1$and $x_2$ are indicated in the figure below.

From the two relations above, we conclude that $$\begin{gather*} k\left(x_1+x_2\right)=\left(m_B+m_p\right)g \\ m_Bg=kx_1\\\\ \Rightarrow m_Bg+kx_2=\left(m_B+m_p\right)g\\\\ \Rightarrow k=\frac{m_pg}{x_2} \end{gather*}$$ Therefore, we have $$k=\frac{0.030\times 9.8}{0.05}=5.88\frac{{\rm N}}{\rm m}$$ 
Frequency of oscillation is $$\begin{align*} f&=\frac{1}{2\pi}\omega =\frac{1}{2\pi}\sqrt{\frac{k}{m_{tot}}}\\\\&=\frac{1}{2\pi}\sqrt{\frac{5.88}{0.030+0.120}}\\\\ &\approx 1\quad {\rm Hz} \end{align*}$$
(b) Traveling from its lowest point to highest point is half of a period. Period of a oscillation is related to the frequency by $T=\frac{1}{f}\approx 1{\rm s}$, therefore the required time is $0.5\,{\rm s}$.
 

(c) From Newton's 2${}^{nd}$ law $F_{net}=ma$, we must find the acceleration of a simple harmonic motion (SHO) for pebble. Recall that in SHO we have the following relations: $$\begin{align*} x=A{\cos  (\omega t+\alpha)\ }\ \ &,\ \ {\rm displacement\ in\ SHO} \\v=\frac{dx}{dt}=-A\omega\,{\sin  \left(\omega t+\alpha\right)\ }\ &,\ \ {\rm velocity} \\ a=\frac{d^2x}{dt^2}=-A\omega^2\,{\cos  \left(\omega t+\alpha\right)\ }\ \ &,\ \ {\rm acceleration}\end{align*}$$ In the relations above, if we substitute the $\left|{\sin  \left(\omega t+a\right)\ }\ or\ {\cos  \left(\omega t+a\right)\ }\right|\le 1$ then we get the maximum values of $x,v$ and $a$. i.e. 
$$x_{max}=A\ ,\ v_{max}=A\omega\ ,\ a_{max}=A\,\omega^2$$
Note: $A$ is the amplitude of the oscillation So 
$$\begin{align*} F_{net}=m_pa_{max}=m\,A\,\omega^2&=mA\sqrt{\frac{k}{m_{tot}}}\\ &=\left(0.030\right)\left(0.12\right)\left(39.2\right)\\&=0.141\quad {\rm N} \end{align*}$$

Now apply Newton's second law to the bob: $\Sigma F=Ma$ $$\begin{align*} F_s+F_g&=M\frac{d^2x}{dt^2}\\\\-Mg\,{\sin \theta}-kx&=M\frac{d^2x}{dt^2}\end{align*}$$  
But from the figure: ${\sin\theta}=x/L$. Therefore $$\begin{align*} -Mg\left(\frac{x}{L}\right)-kx&=M\frac{d^2x}{dt^2}\\\\\Rightarrow \quad -\left(\frac{g}{L}+\frac{k}{M}\right)x&=\frac{d^2x}{dt^2}\end{align*}$$
The above equation is similar to the equation of simple harmonic motion (SHM) i.e. $\frac{d^2x}{dt^2}+\omega^2x=0$, where $\omega$ is the angular frequency of the system.  So 
$$\omega^2=\frac{g}{L}+\frac{k}{M}$$ 
(b) First, find the length of the pendulum. In the absence of the spring $T=2\,{\rm s}$, so 
$$\begin{gather*} \left\{\begin{array}{rcl}\omega=\frac{2\pi}{T} &=& \frac{2\pi}{2}=\pi\\\\ k&=&0 \end{array}\right. \\\\ \Rightarrow \omega=\sqrt{\frac{g}{L}}\Rightarrow L=\frac{g}{\pi^2}\end{gather*}$$
Since the length of the pendulum does not change after placing the spring so substitute $L$ into the angular frequency of the pendulum + spring system and solve for $k$.
$$\begin{gather*} \omega=\frac{2\pi}{1}=\sqrt{\frac{g}{L}+\frac{k}{M}}\\\\ \Rightarrow {\left(2\pi\right)}^2=\pi^2+\frac{k}{M} \\\\ \Rightarrow k=3\pi^2M=3\pi^2\left(1\right)=29.6\ {\rm N/m} \end{gather*}$$

More related articles: 
Simple pendulum Problems and formula
Harmonic motion Problems with formula

(a) The period of an object with mass $m$ under simple harmonic motion with spring constant $k$ is given by 
$$T=\frac{1}{f}=2\pi\sqrt{\frac{m}{k}}$$ 
Where $f$ is the frequency of motion. So $$k=4\,\pi^2f^2m=4\pi^2{\left(2.0\right)}^2\left(0.250\right)=39.5\frac{{\rm N}}{{\rm m}}$$ 

(b) Because there is no friction so mechanical energy is conserved, i.e., it is a constant of motion.

Note: In an SHO, the total mechanical energy is proportional to the square of the amplitude, i.e.,
$$\begin{gather*} E_{tot}=\frac{1}{2}mv^2+U=\frac{1}{2}kA^2 \\\\  =\frac{1}{2}kA^2=\frac{1}{2}(39.5){\left(1.5\times {10}^{-2}\right)}^2=0.0044\,\rm J\end{gather*}$$ For an object at its maximum displacement (amplitude), the total mechanical energy is all the potential energy. Because at this point the velocity of the object is zero.

(c) At the equilibrium position ($x=0$) , the spring is not stretched nor compressed so $U=0$
$$E_{tot}=\frac{1}{2}kA^2=U+K$$ 
$$0.0044=0+\frac{1}{2}mv^2=\frac{1}{2}\left(0.250\right)v^2\Rightarrow v=0.188\ {\rm m/s}$$ 
 

Note: the intensity of a sound wave with average power $P_{av}$ at distance $r$ is given by 
$$I=\frac{P_{av}}{4pr^2}$$ 
Solving for $r$, we obtain
$$\Rightarrow r=\sqrt{\frac{P_{av}}{4\pi I}}=\sqrt{\frac{12}{4\pi\times {10}^{-12}}}=9.77\times {10}^5\ {\rm m}$$ 

(a) Recall that the wave speed on the sting with tension $T$ and linear mass density $\mu$ is given by 
$$v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{5\times {10}^3}{8\times {10}^{-2}}\ }=250\frac{{\rm m}}{{\rm s}}$$ 

(b) For a string fixed at both ends, the standing wave condition is $L=n\frac{\lambda_n}{2}\$and using the relation $v=\lambda_n f_n$ we can find the frequency of the nth harmonic as $f_n=n\frac{v}{2L}$. So 
$$f_n=n\frac{v}{2L}\ ,\ n=1,2,3,\dots \ \ but\ \ n=1\ is\ fundamental\ mode$$ 
$$f_1=\left(1\right)\frac{v}{2L}=\left(1\right)\frac{250}{2\left(0.5\right)}=250\ {\rm Hz\ \ \ ,\ }{\lambda }_{{\rm 1}}=\frac{v}{f_1}=\frac{250}{250}=1\ {\rm m}$$ 

(c) Recall that when a wave passing through a medium; its frequency does not change but the wavelength changes. So 
$$\lambda_1=\frac{v_{sound}}{f_1}=\frac{343}{250}=1.372\ {\rm m}$$ 
That is, the wavelength of the fundamental mode in air is greater than the string.
 

As mentioned previously, the frequency of the nth harmonic in an excited string with both ends fixed is given by the $f_n=n\frac{v}{2L}$. Where $v$ is speed of the wave on the string with tension $T$ i.e. $v=\sqrt{\frac{T}{\mu}}$. Therefore, first find the speed of the wave 
$$v=\sqrt{\frac{T}{\frac{m}{L}}}=\sqrt{\frac{600}{\frac{0.3\times {10}^{-3}\ {\rm kg}}{{\rm 0.7\ m}}}}=1183.2\frac{{\rm m}}{{\rm s}}$$ 
$$f_1=\left(1\right)\frac{1183.2}{2\left(0.7\right)}=845\ {\rm Hz\ }\ \ {\rm fundamental\ harmonic\ }\left({\rm n=1}\right)$$ 
$$f_2=\left(2\right)\frac{1183.2}{2\left(0.7\right)}=1690\ {\rm Hz\ \ \ }{{\rm 2}}^{{\rm nd}}{\rm \ harmonic}$$ 
$$f_3=\left(3\right)\frac{1183.2}{2\left(0.7\right)}=2535\ {\rm Hz\ }\ {{\rm 3}}^{{\rm nd}}{\rm \ harmonic}$$