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 
\[v=\sqrt{\frac{T}{\frac{m}{L}}}=\sqrt{\frac{600}{\frac{0.3\times {10}^{-3}\ {\rm kg}}{{\rm 0.7\ m}}}}=1183.2\frac{{\rm m}}{{\rm s}}\] 
\[f_1=\left(1\right)\frac{1183.2}{2\left(0.7\right)}=845\ {\rm Hz\ }\ \ {\rm fundamental\ harmonic\ }\left({\rm n=1}\right)\] 
\[f_2=\left(2\right)\frac{1183.2}{2\left(0.7\right)}=1690\ {\rm Hz\ \ \ }{{\rm 2}}^{{\rm nd}}{\rm \ harmonic}\] 
\[f_3=\left(3\right)\frac{1183.2}{2\left(0.7\right)}=2535\ {\rm Hz\ }\ {{\rm 3}}^{{\rm nd}}{\rm \ harmonic}\]

Note: the sound intensity levels are measured on logarithmic scale as $\beta=10\, {\rm dB}{\log  \frac{I}{I_0}\ }$, where $I_0={10}^{-12}\ {\rm W/}{{\rm m}}^{{\rm 2}}$ is the threshold of hearing. First, find the intensity of one fly 
\[\beta=10\ {\rm dB}\,{\log  \frac{I_0}{I}\ }\Rightarrow 40=10\,{\log  \frac{I}{I_0}\ }\Rightarrow \frac{I}{I_0}={10}^4\Rightarrow I_{one}={10}^4I_0\] 
The intensity of 1000 flies is 1000 times the intensity of one fly that is $I_{1000}=1000I_{one}={10}^7I_0$
\[\beta_{1000}=10\ {\rm dB}\,{\log  \frac{I_{1000}}{I_0}\ }=10\,{\log  \frac{{10}^7I_0}{I_0}\ }\Rightarrow \ \beta_{1000}=70\ {\rm dB}\] 

(a) If on end of a tube with length $L$ is fixed (closed), then the condition of the standing waves states that the frequency of nth harmonic is $f_n=nv/4L$. So 
\[f_1=\left(1\right)\frac{343}{4\left(0.240\right)}=357.29\, {\rm Hz}\] 

(b) The speed of sound waves in the gasses is given by $v=\sqrt{\frac{\gamma kT}{m}}$, where $T$ is the absolute temperature measured in Kelvins (${\rm K}$) and $m$ is the mass of a molecule of the gas. Therefore, first find the speed of the sound in the helium then calculate the fundamental frequency in it.
\[v=\sqrt{\frac{\gamma RT}{M}}=\sqrt{(1.4)(\frac{\left(1.38\times {10}^{-23}\right)\left(273+37{\rm{}^\circ\!\,C}\right)}{3.32\times {10}^{-27}}}=1342.12\frac{{\rm m}}{{\rm s}}\] 
\[f^{'}_1=n\ \frac{v}{4L}=\left(1\right)\frac{1342.12}{4\left(0.240\right)}=1399.1{\rm Hz}\ \] 
 

(a) If both ends of a string with length $L$ are fixed, then the frequency of the nth harmonic wave is determined by $f_n=n\frac{v}{2L}$, where $v$ is the speed of the wave on the string with tension $T$ and is found by $v=\sqrt{\frac{T}{\frac{m}{L}}}$. Therefore,
\begin{align*} 2L f_n&=n\sqrt{\frac{T}{\frac{m}{L}}} \\\\ \Rightarrow T&=\frac{4Lf^2_nm}{n^2}\\\\&=\frac{4(0.35)(440)^2 (0.5\times {10}^{-3})}{1^2}\\\\&=136\quad {\rm N}\end{align*}
(b) First draw a vector from the receiver toward the source then choose the signs of the Doppler formula as follows (choose plus if any velocity is in the direction of this vector and vice versa).
\begin{align*}f_r&=f_s\frac{v\pm v_r}{v\pm v_s}\\\\&=f_s\frac{v+v_r}{v}\\\\&=440\frac{343+20}{343}\\\\&=465.6\quad {\rm Hz} \end{align*}
(c) The intensity level of the sound is measured as $\beta=10\,{\log \frac{I}{I_0}}$, where $I$ is the intensity of the source of the sound at distance $r$ from it and is given by $I=\frac{P}{4\pi r^2}$ so by combining these equations we get
\begin{gather*}\beta=10\,{\log \frac{P}{{4\pi r^2I}_0}}\\\\ \Rightarrow \frac{P}{{4\pi r^2I}_0}={10}^{\frac{\beta}{10}}\\\\ \Rightarrow P=4\pi r^2I_0\times {10}^{\frac{\beta}{10}}\end{gather*}
Where $I_0={10}^{-12}\, {\rm W/m^{2}}$ is the threshold of hearing. Therefore,
\begin{align*} P&=4\pi\times 5^2\times {10}^{-12}\times {10}^{\frac{60}{10}}\\\\&=314.15\times {10}^{-6}\quad {\rm W}\\\\&=314.15\quad {\rm \mu W}\end{align*}

(a) The intensity of 20 violins is $20$ times the intensity of one violin, i.e., $I_{20}=20\,I_1$, so use the relation of the intensity level to find $I_{20}$
\begin{gather*} \beta_{20}=10\,{\log  \frac{I_{20}}{I_0}\ }\\ \\ \Rightarrow 82.5=10\,{\log\left(\frac{I_{20}}{I_0}\right)}\\\\\Rightarrow I_{20}={10}^{8.25}I_0\end{gather*} Hence, \[I_{20}=20\,I_1\Rightarrow I_1=\frac{{10}^{8.25}}{20}I_0\] 
Now compute the intensity level for one violin as 
\begin{gather*} \beta_1=10\,{\log  \frac{I_1}{I_0}}\\ \\ \to \beta_1=10\,{\log  \left(\frac{{10}^{8.25}}{20}\right)}=69.48\quad {\rm dB}\end{gather*}
(b) As before, $I_{40}=40I_1$. Therefore, \begin{align*} I_{40}&=10\,{\log  \left(\frac{I_{40}}{I_0}\right)}\\\\&=10\,{\log  \left(\frac{40\times\frac{{10}^{8.25}}{20}I_0}{I_0}\right)}\\\\&=85.5\ {\rm dB<165\ dB}\end{align*} You can find more problems on sound intensity level here. 

(a) The intensity level ($\beta$) of the sound waves is measured in ${\rm dB}$ by the following equation
\[\beta=10\,{\log  \frac{I}{I_0}\ }\] 
Where $I$ is the intensity of the source of sound at distance $r$ and $I_0$ is the hearing threshold. So find the ratio $I_2/I_1$ as follows
\[\beta_1=10\,{\log  \frac{I_1}{I_0}\ }\Rightarrow 20=10\,{\log  \frac{I_1}{I_0}\ }\Rightarrow \frac{I_1}{I_0}={10}^2\] 
\[\beta_2=10\,{\log  \frac{I_2}{I_0}\ }\Rightarrow 60=10\,{\log  \frac{I_2}{I_0}\ }\Rightarrow \frac{I_2}{I_0}={10}^6\] 
We have used the definition of the logarithm i.e. $y={\log  x\ }\Rightarrow x={10}^y$. Dividing the two relation, we obtain 
\[\frac{I_2}{I_1}=\frac{{10}^6}{{10}^2}={10}^4\] 

(b) The sound intensity $I$ is defined as the sound power $P$ that passes perpendicularly through a surface divided by the area $A$ of that surface $I=P/A$. For a point source at distance $r$ it is $I=P/4\pi r^2$ so 
\[I_1=\frac{P}{4\pi r^2_1}\ \ and\ \ I_2=\frac{P}{4\pi r^2_2}\] 
Dividing the two equation, gives 
\[\frac{r_1}{r_2}={\left(\frac{I_2}{I_1}\right)}^{\frac{1}{2}}={\left({10}^4\right)}^{\frac{1}{2}}=100\ \] 
Therefore, $$\frac{r_2}{r_1}=\frac{1}{100}=0.01$$
 

(a) If the both ends of a tube is open then the frequency of the $n$th harmonic is given by the $f_n=n\frac{v}{2L}\ ,\ n=1,2,3,\dots $ . Consider the $nt$h frequency is $440\, {\rm Hz}$, the next frequency is $\left(n+1\right)$th. Therefore,
\[f_n=n\frac{v}{2L}=440\ {\rm Hz\ ,\ \ }f_{n+1}=\left(n+1\right)\frac{v}{2L}=528\ {\rm Hz}\] 
Subtracting them, we obtain 
\[f_{n+1}-f_n=528-440\to \frac{v}{2L}=88\] 
Now use the above result to find the fundamental frequency $f_1=\left(1\right)\frac{v}{2L}=88\ {\rm Hz}$

(b) Let the speed of the sound be $v=343\, {\rm m/s\ }$, therefore, 
\[\frac{v}{2L}=88\Rightarrow l=\frac{343}{2\left(88\right)}=1.94\ {\rm m}\] 
 

(a) The speed of the waves on a string with tension $T$ and mass density $\mu=m/L$ is given by
\[v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{100}{\frac{0.007}{0.45}}}=80.17\, \frac{{\rm m}}{{\rm s}}\] 
(b) Since both ends of a guitar string is fixed so the frequency of the nth harmonic is found by
\[f_n=n\frac{v}{2L}=\left(3\right)\frac{80.17}{2\left(0.45\right)}=267.23\ {\rm Hz}\] 
(c) First, compute the fundamental frequency of the string and associated wavelength in the air
\[f_1=\left(1\right)\frac{v}{2L}=\left(1\right)\frac{80.17}{2\left(0.45\right)}=89.07\ {\rm Hz}\] 
\[\lambda_1=\frac{v_{sound}}{f_1}=\frac{343}{89.07}=3.85\ {\rm m}\] 
Listener hears no sound means that at the location of listener the waves must be destructive! Therefore, use the destructive interference pattern in waves that is the path difference of waves must be equal the half integer of wavelength i.e. $\Delta =n\lambda\ ,\ \ n=\frac{1}{2},\frac{3}{2},\dots $
\[\Delta =x-d=n\lambda \Rightarrow x=\left(d+n\lambda \right)\]
squaring both sides, we obtain
\[x^2={\left(d+n\lambda\right)}^2=d^2+n^2\lambda^2+2n\lambda d\] 
Now as shown in the figure, use the Pythagorean theorem to write $x^2=d^2+4^2$. Substitute this into the relation above
\[d^2+16=d^2+{\left(n\lambda\right)}^2+2\left(n\lambda\right)d\Rightarrow d=\frac{16-{\left(n\lambda\right)}^2}{2\left(n\lambda\right)}\] 
\[d=\frac{16-{\left(\frac{1}{2}\lambda\right)}^2}{2\left(\frac{1}{2}\lambda\right)}=\frac{16-\frac{{\left(3.85\right)}^2}{4}}{3.85}=3.19\ {\rm m}\] 
If $n=3/2\ $then $d=1.50$ so the largest value of $d$ is $3.19\ {\rm m}$.

Apply the Doppler shift equation as
\[f_r=f_s\frac{v\pm u_r}{v\pm u_s}\] 
Where $f_r$ and $f_s$ are the frequencies of the receiver and source and also $u_r\ ,\ u_s$ are the receiver and source velocities, respectively. 
Important note about signs in the Doppler shift: First, suppose a direction from receiver toward the source then all velocities in the direction of this vector are positive and all velocities in opposite direction are negative.   
In this problem car is the receiver and train is the source of sound. Using these notes, we have: $v=343\, {\rm m/s}$ , $f_s=1500\, {\rm Hz\ ,\ }u_r=9.5\, {\rm m/s}$ , $u_s=22.5\, {\rm m/s}$ ,  $f_r=?$  
(a) The figure below shows the situation:
\[f_r=f_s\frac{v-u_r}{v-u_s}=1500\frac{343-9.5}{343-22.5}=1560.8\ {\rm Hz}\] 

(b) Our sketch for this case is as follows:
\[f_r=f_S\frac{v+u_r}{v+u_s}=1500\frac{343+9.5}{343+22.5}=1446.6\ {\rm Hz}\] 

As shown in the figure, there are two velocities with respect to the listener. First, use the relation $v=r\omega$ to find the velocity of the source then apply Doppler Effect to calculate the frequencies associated with these sources.

\[v_s=r\omega=5\times 0.8=4\frac{{\rm m}}{{\rm s}}\] 
To use the Doppler Effect, as previously mentioned, draw a vector from the listener to the source then compare the signs of velocities in the Doppler equation with this vector.
\[f_o=f_s\frac{v\pm u_o}{v\pm u_s}\] 
\begin{align*} \text{Case one:} \quad f_o&=f_s \left(\frac{v}{v+u_s}\right) \\\\ &=750\left(\frac{343}{343+4}\right) \\\\ &=741.35\, {\rm Hz} \\\\  \text{Case two:} \qquad \ f_o&=f_s\left(\frac{v}{v-u_s}\right)\\\\ &=750\left(\frac{343}{343-4}\right)\\\\ &=758.84\ {\rm Hz} \end{align*}
Now with the given frequencies find the associated wavelengths of them as
Case one: \[\lambda_1=\frac{v}{f_o}=\frac{343}{741.35}=0.462\, {\rm m}=46.2\ {\rm cm}\]
Case two: \[\lambda_2=\frac{v}{f^{'}_o}=\frac{343}{758.84}=0.452\, {\rm m}=45.2\ {\rm cm}\]
So the longest wavelength is $46.2\, {\rm cm}$. 

(a) Use the Doppler Effect. To do this First draw a vector from the receiver toward the source and then find the signs of the velocities with respect to this vector. 
\[f_r=f_s\frac{v\pm u_r}{v\pm u_s}=f_s\frac{v}{v-u_s}=500\frac{343}{343-25}=539.3\ {\rm Hz}\] 

(b) In this case, the mountain is as source with the reflected frequency $539.6\, {\rm Hz\ }$and the train conductor is receiver. So the frequency that conductor receives is 
\[f_{train}=f_{moun}\frac{v\pm u_{train}}{v\pm u_{moun}}=f_{moun}\frac{v+u_{train}}{v}=539.3\frac{343+25}{343}=578.6\ {\rm Hz}\] 

(c) By definition, beat frequency is the difference of two frequencies: 
\[f_{beat}=\Delta f=f_{train}-f_{whistle}=978.6-500=78.6\ {\rm Hz}\] 
 

Use the Doppler Effect for each of them. The received frequency by the eagle $1$ is \begin{align*} f_{O_1}&=f_{S_2}\frac{v\pm u_{O_1}}{v\pm u_{S_2}} \\\\ &=(3800) \frac{330+15}{330-20}\\\\ &=4229\quad \rm Hz\end{align*} In the case of eagle $2$, the eagle $1$ is a source and eagle $2$ is a receiver (or object). Thus, the frequency perceived by the eagle $2$ is found to be \begin{align*} f_{O_2}&=f_{S_1}\frac{v\pm u_{O_2}}{v\pm u_{S_1}}\\\\ &=(3200) \frac{330+20}{330-15} \\\\ &=3556\quad \rm Hz\end{align*} Practice more on the Doppler effect problems.

a) Here ambulance is the source and the standing person is the receiver (object) since $f_o=0.9\,f_s$, so the ambulance moves away from the observer (you). 

b) Use the Doppler Effect to find the speed of the ambulance.
\begin{gather*} f_s=f_o\frac{v\pm u_s}{v\pm u_o}\\\\ \Rightarrow f_s=f_o\ \left(\frac{v+u_s}{v}\right) \\\\ \Rightarrow \ f_s=0.9f_s\,\left(\frac{343+u_s}{343\ }\right) \\\\ \Rightarrow \boxed{u_s=38.11\, {\rm m/s}}\end{gather*} 

c) Given data: $u_o=3\, {\rm m/s}$, $u_s=38.11\, {\rm m/s}$, $f_s=900\, {\rm Hz}$ , $f_o=?$ As shown in the figure we have
\begin{align*} f_o&=f_s\frac{v\pm u_o}{v\pm u_s} \\\\ &=f_s\frac{v-u_o}{v+u_s} \\\\ \Rightarrow f_o &=900\frac{343-3}{343+38.11} \\\\ &=802.9\ {\rm Hz}\end{align*}

(a) The wavelength emitted from a source is related to the frequency by
\[\lambda=\frac{v}{f}=\frac{1482}{22\times {10}^3}=0.0673\,{\rm m}\] 
(b) In first case, the sonar is source and the whale is receiver (object) so the frequency that the whale perceives is:
\[f_w=f_s\frac{v+u_w}{v+u_s}=\left(22\times {10}^3\right)\left(\frac{1482+4.95}{1482+0}\right)=22073.4\,{\rm Hz} = 22.073\,{\rm kHz}\] 

In second case, the reflected frequency from whale (in first case) is source and the sonar is receiver (object). Thus the frequency that the sonar received is 
\[f_s=f_w\frac{v+u_s}{v+u_w}=22.073\ \left(\frac{1482}{1482-4.95}\right)=22.17\times {10}^3\ {\rm Hz}\]

 
\[\Delta f=22.17-22.073=0.097{\rm kHz}\] 

(a) The speed of waves in fluids such as air or water is given by $v=\sqrt{\frac{B}{\rho}}$ where B is the Bulk modulus and $\rho$ is the density of the medium. Therefore 
\[v=\sqrt{\frac{B}{\rho}}=\sqrt{\frac{2.2\times {10}^9}{1000}}=1483\, {\rm m/s}\] 
(b) Having the sound speed in the water $v$, we can find its wavelength by
\[\lambda=\frac{v}{f}=\frac{1483}{1200}=1.24\ {\rm m}\] 
(c) Fluids exerts a drag force on small, slowly moving spheres by stock's law as $F_d=6\pi \eta rv$. In addition to the drag force, there is also a buoyancy force. Thus, draw a free body diagram and apply Newton's second law as follows \[\Sigma F_y=ma=0\ \left(const\ velocity\right)\] 
\[mg-F_B-F_D=0\] 
\[mg=\rho_w\, V_{submerg}\,g+6\pi\eta rv\] Where $F_B$ is the buoyant force exerted on the body from the fluid surrounding it whose magnitude is obtained using the well-known buoyant force formula.

Solving the above equation for $v$, we have \begin{align*} v&=\frac{mg-\frac{4}{3}\pi r^3\rho_w g}{6\pi\eta r}\\ &=\frac{\left(0.015\times {10}^{-3}\times 9.8-\frac{4}{3}\pi\times {\left(1.1\times {10}^{-3}\right)}^3\left(1000\right)\left(9.8\right)\right)}{2.07\times {10}^{-3}}\\ &=4.46\frac{{\rm m}}{{\rm s}} \end{align*} Now suppose this sinking speaker as a source with velocity $v_s=4.46\, {\rm m/s}$ and the stationary speaker as the receiver (object). Applying Doppler effect and solving for the frequency received by the stationary speaker, we obtain \begin{align*} f_r&=f_s\frac{v\pm u_r}{v\pm u_s}\\\\&=1200\frac{1483+0}{1483+4.46}\\\\&=1196.4\quad {\rm Hz}\end{align*}
This is a frequency that the stationary speaker receives. Recall from the beats frequency problems section, the frequency of beats heard is \[\left|\Delta f\right|=\left|f_r-f\right|=\left|1196.4-1200\right|=3.6\, {\rm Hz}\] 

First draw a vector from receiver (bicyclist) to the source of sound (car) and then by using it determine the signs in the Doppler formula as follows
\[f_r=f_s\frac{v\pm v_r}{v\pm v_s}\] 

\[f_r=f_s\frac{v+v_r}{v+v_s}\] 
Given data are: $f_s=480\ {\rm Hz}$ , $v_r=\frac{1}{3}v_s$ , $f_r=422\ {\rm Hz}$
\[\Rightarrow 422=480\frac{345+\frac{1}{3}v_s}{345+v_s}\] 
\[v_s=76.37\,\rm m/s\] 

The waves that are produced in a string with both ends fixed are called standing waves. In this case, the length of the string is equal to some integer multiple of half-wavelengths i.e. $L=n\ \lambda /2$, or its wavelengths are quantized as ${\lambda }_n=2L/n$ where ${\lambda }_n$ is the wavelength of the $n$th mode.

In the other words, a standing wave can exist only if its wavelength satisfies the equation above. The fundamental relation between frequency $f$ and wavelength $\lambda $ of a traveling wave with velocity $v$ is $v=f\lambda $. Therefore, the frequency of $n$th mode is found using the standing wave formula below \[f_n=n\frac{v}{2L}\quad ,\quad \left(n=1,2,3,\dots \right)\] 
The frequency of $n=1$ case is called the fundamental frequency i.e. $f_1=\frac{v}{2L}$. 
The $50\,{\rm Hz}$ frequency in above is fundamental frequency so using it we can find the ratio of $v/L$ as 
\begin{align*} f_1&=\frac{v}{2L}\\\\ 50&=\frac{v}{2L}\\\\\Rightarrow \frac{v}{L}&=100 \end{align*}
Note: The speed of waves in a medium depends on the physical properties of that so in the second stage the speed of waves $v$ no longer changes. 
At the second stage, we have a standing wave again with frequencies as $f^{'}_n=n\ v/2L^{'}$. In this case $L^{'}=L/2$. Thus
\[n=1\to f_1=\frac{v}{2L^{'}}=\frac{1}{2}\left(\frac{v}{\frac{L}{2}}\right)=\frac{v}{L}=100\ \mathrm{Hz}\] 
\[n=2\to f_2=2\frac{v}{2L^{'}}=\frac{v}{\frac{L}{2}}=2\left(100\right)=200\ \mathrm{Hz}\] 
\[n=3\to f_3=3\frac{v}{2L^{'}}=\frac{3}{2}\frac{v}{\frac{L}{2}}=3\left(100\right)=300\ \mathrm{Hz}\] 
\[n=4\to f_4=4\frac{v}{2L^{'}}=2\frac{v}{\frac{L}{2}}=4\ \left(100\right)=400\ \mathrm{Hz}\] 

The correct answer is C.

By definition of the intensity of waves as the ratio of the time rate of energy transported to the unit area we have 
\[I=\frac{P}{A}\] 
Where $P$  which is the time rate of energy transported is called power of the source. So \[I=\frac{1.1\times {10}^{-3}\,\rm W}{2\times 10^{-4}\,\rm m^2}=5.5\,\rm W/m^2\]

The correct answer is E. 

As mentioned earlier, the normal mode frequencies of strings at different conditions are given by the following equations
\[f_n=n\frac{v}{2L}\left(n=1,2,3,\dots \right)\ \ \ \ \ both\ ends\ fixed\ or\ open\] 
\[f_n=n\frac{v}{4L}\left(n=1,3,5,\dots \right)\ \ \ \ \ open\ at\ only\ one\ end\] 
In A the length of pipe is $L=2.3\ \mathrm{m}$ and fixed at one end so its normal modes are
\[f_n=n\frac{v}{4L}\to \ \ n=1\ ,\ \ f_1=\frac{v}{4\left(2.3\right)}=\frac{v}{9.2}\] 
In B, $L=3.3\ \mathrm{m}$ and its conditions is same as the $A$
\[f_n=n\frac{v}{4L}\to \ \ n=1\ ,\ \ f_1=\frac{v}{4\left(3.3\right)}=\frac{v}{13.2}\] 
In C, $L=1.6\ \mathrm{m}$ and closed at both ends, 
\[f_n=n\frac{v}{2L}\to \ \ n=1\ \ ,\ \ f_1=\frac{v}{2\left(1.6\right)}=\frac{v}{3.2}\] 
In D, $L=3.0\ \mathrm{m}$
\[f_n=n\frac{v}{2L}\to \ \ n=1\ \ ,\ \ f_1=\frac{v}{2\left(3.0\right)}=\frac{v}{6}\] 
In E, displacement nodes are $5\ \mathrm{m}$ apart that is the fundamental wavelength of such a pipe is ${\lambda }_1=5\ \mathrm{m}$, using the relation between speed and frequency $f=\frac{v}{\lambda }$, we have
\[f_1=\frac{v}{5}\] 
Since the speed of waves on a string depends on its linear mass density and tension on it therefore 

The correct answer is C.
 

In a circular motion there always is a centripetal force toward the center of the circle that in this case is provided by the Coulomb's force between the charges. Using Newton's second law of motion along the radius, we get
\[\Sigma\vec{F}=m\vec{a} \Rightarrow F_e=\frac{mv^2}{r}\] where $F_e$ is Coulomb's force between two charged particles. 
\begin{gather*} k\frac{\left|q_1q_2\right|}{r^2}=m\frac{v^2}{r}\\\\ \Rightarrow r=k\frac{\left|q_1q_2\right|}{mv^2}\\\\ =\left(9\times {10}^9\right)\frac{{\left(5\times {10}^{-6}\right)}^2}{\left(20\times {10}^{-3}\right){\left(7\right)}^2}=0.23\quad {\rm m}\end{gather*}

The correct answer is B. 

In Young's double-slit experiment, the distance from the centerline to the center of a fringe is given by $y=D\ m\lambda /d$ where $r,m,\lambda $ and $d$ are the screen-to-slit separation, $m$th bright or dark fringe, $\lambda $ the wavelength of the source and $d$ is the distance between slits, respectively. 
In the case of bright fringes $m=0,\pm 1,\pm 2,\dots $ and for dark fringes $m=\pm \frac{1}{2},\pm \frac{3}{2},\dots $ . 
The distance between two fringe is found by two consecutive $m$th as follows
\begin{align*} \Delta &y=y_{m+1}-y_m \\\\ &=D\frac{(m+1)\lambda}{d}-D\frac{m\lambda}{d} \\\\ &=D\frac{\lambda}{d}\end{align*} 
Because the source does not change $\lambda =\lambda'$ so 
\begin{gather*} \frac{\Delta y}{\Delta y'}=\frac{D}{D'}\frac{d'}{d} \\\\  1=\frac{D}{D'}\frac{2d}{d} \\\\ \Rightarrow D'=2D \end{gather*}

The correct answer is D.