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A pipe closed at both ends can support standing waves, but the sound cannot get out unless singing in the shower can be compared with singing in a closed pipe.

(a) Show that the wavelength of standing waves in a closed pipe of length $L$ are $\lambda_n=2L/n$ and the frequencies are given by $f_n=\frac{nv}{2L}=nf_0$ where $n=1,2,3,\dots $

(b) Modeling the shower as a closed pipe, find the frequency of the fundamental and the first two overtones if the shower is 2.5 m tall. Are these frequencies audible?

(c) You remembered the outside showers facilities at summer-camp and how your voice sounded different, and you thought-maybe outside camp showers were stopped pipes instead closed pipes-no ceiling! Repeat your calculations of part (a), and (b) and note the difference in the two situations

(a) Since both ends of the string is fixed, so 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.

Suppose we have massive rope attached to the ceiling. The bottom end of the rope dangles in midair. The mass of the rope is $2.7\ {\rm kg}$ and its length is $4.5\ {\rm m}$. The bottom end of the rope is shaken and then released to produce a pulse that travels up the rope. What is the speed of the pulse when it is halfway up the rope?

(a) The figure shows a snapshot graph at $t=0$ of a sinusoidal wave travelling to the right along a string at $50\ {\rm m/s}$. Write the equation that describe the displacement $D(x,t)$ of this wave. Your equation should have numerical values, including units, for all quantities except $x$ and $t$.

(b) What is the maximum acceleration of any portion of the string described above?

A standing wave is oscillating at $690\ {\rm Hz}$ on a string of mass $1{\rm g}$, as shown in the figure. What is the tension in the string?

\[L=3\frac{\lambda}{2}\Rightarrow \ \lambda=\frac{2}{3}L\]

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*}

Two identical loudspeakers are located at points A and B, 2 meters apart. The speakers are driven by the same sound source with a frequency of $784\ {\rm Hz}$. Take the speed of sound to be $344\ {\rm m/s}$. A small microphone is moved out from the point $B$ toward the point $C$ a shown.

(a) At what distance from $B$ will there be destructive interference?

(b) At what distance from $B$ will there be instructive interference?

(c) If the frequency is made low enough, there will be no position along the line $BC$ at which destructive interference can occur. How low must the frequency be for this to be the case?

Monochromatic ($\lambda=633\,{\rm nm}$), coherent plane waves of light from a laser are incident on two slits that are separated by a distance $d=0.070\,{\rm nm}$. Independently (without interference effects), each slit illuminates the screen $3.00\ {\rm m}$ away with an intensity of $200\ {\rm W/}{{\rm m}}^{{\rm 2}}$ at the screen.

(a) Determine the positions $y_{min}(m)$ (for $y_{min}>0$) on the screen of the first three intensity minima, where m is the order of each minimum? What is the intensity $I_{min}$ at these positions?

(b) Determine the positions $y_{max}$ (for $y_{max}>0$)on the screen of the first three intensity maxima, where m is the order of each maximum? What is the intensity $I_{max}$ at these positions?

(c) Determine the position $y_{1/4\ }$at which the intensity of the central peak (at $y=0$) drops to $I_{max}/4$. Hint: calculate the angles in radians.

The picture below shows unpolarized light incident from the left upon a pair of polarizes. The intensity of the unpolarized light is $15\ {\rm W/}{{\rm m}}^{{\rm 2}}$. The vertical dashed line in the left polarizer shows its transmission axis (TA).

(a) What is the intensity of the light as it passes point A?

(b) After passing point A, the light now passes through a second polarizer, and then passes point B. At B the light's intensity is $3\ {\rm W/}{{\rm m}}^{{\rm 2}}$. What is the smallest angle between the directions of the TA's of the two polarizers?

In a Young's double slit experiment ($d=0.2\,{\rm mm}$) coherent light from two sources $\lambda_1=450\,{\rm nm}$ and $\lambda_2=600\,{\rm nm}$ illuminate the double slit (see figure). There is a point P on the screen, a distance $y$ up from the center line at which a maximum from $\lambda_1$ overlaps a minimum from $\lambda_2$. This point P is the closest such point to the centerline for the described overlapping.

(a) What is the position $y$ at which this overlapping occur?

(b) Counting the central bright fringe as the first bright fringe ($0$${}^{th}$ order bright fringe) and the first dark fringe above the center line ($0$${}^{th}$ order dark fringe) as the first dark fringe, then at the position P which bright fringe is overlapping which dark fringe?

Two thin parallel slits that are $0.0116\,{\rm mm}$ apart are illuminated by a laser beam of wavelength $585\,{\rm nm}$. The resulting interference pattern is presented on a very large distant screen.

(a) What is the largest value of ${\sin \theta\ }$ that we have to consider?

(b) What is the total number of bright fringes that we have to consider?

(c) At what angle, relative to the original beam direction will most distant fringe appear?

Unpolarized light of intensity $S_0$ is incident from the left on a series of polarizes. The angle between the transmission axes of $P_1$ and $P_2$ is $60{}^\circ $.

(a) In terms of $S_0$, the average intensities at $A$ and $B$ are:

(b) A third polarizer $P_3$ is placed first at $A$ then at $B$.

(i) What must be the angle between the transmission axes of $P_1$ and $P_3$ so that no light reaches when $P_3$ is placed at $A$?

(ii) What must be the angle between the transmission axes of $P_1$ and $P_3$ so that no light emerges to the right of $P_3$ when $P_3$ is placed at $B$?

(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 in B. light passes through two polarizer before incident on it so

\[S_B=\frac{S_0}{8}\,{{\cos }^2 \phi^{''}\ }=0\left(no\ light\ transmitted\right)\Rightarrow \ \phi^{''}=90{}^\circ \]

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 $.

A suitcase of mass $20\, {\rm kg}$ is hung from two bungee cords. Each cord is stretched $5\, {\rm cm}$ when the suitcase is in equilibrium. If the suitcase is pulled down a little and released, what will be its oscillation frequency?

A $0.120\,{\rm kg}$ block is suspended from a spring. When a small pebble of mass $30\,{\rm g}$ is placed on the block, the spring stretches an additional $5.0\,{\rm cm}$. With the pebble on the block, the block oscillates with an amplitude of $12\,{\rm cm}$.

(a) What is the frequency of the motion?

(b) How long does the block take to travel from its lowest point to its highest point?

(c) What is the net force on the pebble when it is at the point of maximum upward displacement?

Figure shows a pendulum of length $L$ with a bob of mass $M$. The bob is attached to a spring that has a force constant $k$. When the bob is directly below the pendulum support, the spring is unstressed.

(a) Derive an expression for the period of this oscillating system for small amplitude vibrations.

(b) Suppose that $M=1.00\,{\rm kg}$ and $L$ is such that in the absence of the spring the period is $2.00\,{\rm s}$. What is the force constant $k$ if the period of the oscillating system is $1.00\,{\rm s}$?

A child's spring-toy consists of a $250\ {\rm g}$ plastic frog which bounces up and down on a spring. When the frog is pushed down by $1.5\,{\rm cm}$ below its equilibrium position and released, it vibrates up and down at a frequency of $2.0$ times per second.

(a) Find the spring constant $k$ of the spring.

(b) Assuming no friction, what is the total energy of the bouncing frog?

(c) Find the velocity of the frog when it passes through equilibrium position (where $x=0$)

(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. is a constant of motion.

Note: In a SHO, the total mechanical energy is proportional to the square of amplitude i.e.

\[E_{tot}=\frac{1}{2}mv^2+U=\frac{1}{2}kA^2\]

\[\therefore \ E_{tot}=\frac{1}{2}kA^2=\frac{1}{2}\left(39.5\right){\left(1.5\times {10}^{-2}\right)}^2=0.0044{\rm \ J}\]

For an object at its maximum displacement (amplitude), the total mechanical energy is all the potential energy. Because at this point the velocity of object is zero.

(c) At the equilibrium position ($x=0$) , the spring is not stretched nor compressed so $U=0$

\[E_{tot}=\frac{1}{2}kA^2=U+K\]

\[0.0044=0+\frac{1}{2}mv^2=\frac{1}{2}\left(0.250\right)v^2\Rightarrow v=0.188\ {\rm m/s}\]

A small sound source has a power of $12\,{\rm W}$. How far away from the sound source can you be and still hear it? Recall the threshold of hearing is $1.00\times {10}^{-12}\ {\rm W/}{{\rm m}}^{{\rm 2}}$.

A guitar string of length $0.5\, {\rm m}$ and linear mass density of ${\rho }_{string}=8.00\times 10^{-2}\,{\rm \ kg/m}$ is held with a tension of ${\rm 5.00\ \times \ 103\ N}$. It is plucked so as to excite the fundamental frequency alone

(a) What is the velocity of a wave on the string?

(b) What is the wavelength and frequency of the fundamental harmonic?

(c) What is the wavelength of the sound in air?

A $0.3\, {\rm g}$ wire is stretched between two points $70\, {\rm cm}$ apart. If the tension in the wire is $600\, {\rm N}$ find the wire's first, second and third harmonics?

If a large horsefly $3.0\, {\rm m}$ away from you makes a noise of $40.0\, {\rm dB}$, what is the noise level of $1000$ flies at that distance?

A crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end.

(a) What is the fundamental frequency if the tube is $0.240\, {\rm m}$ long?

(b) What would the frequency become if the person replaced the air with helium?

($k=1.38\times {10}^{-23}\ {\rm J/K}$, $m_{He}=3.32\times {10}^{-27}{\rm kg}$, air temperature $37{\rm{}^\circ\!\,C}$ , $\gamma=1.4$)

A violin string with a mass of $0.5\, {\rm g}$ and a length of $35\, {\rm cm}$ produces a fundamental frequency of $440\, {\rm Hz}$ when bowed.

(a) What is the tension in the string?

(b) If an observer moves toward the violin at a speed of $20\, {\rm m/s}$, what frequency is heard?

(c) If the sound intensity is $60\, {\rm dB}$ at a distance of $5\, {\rm m}$ from the violin, what acoustical power does the violin produce, assuming it radiates equally in all directions?

Twenty violins playing simultaneously with the same intensity combine to give an intensity level of $82.5\, {\rm dB}$.

a) What is the intensity level of each violin?

b) If the number of the violins is increased to 40, will the combined intensity level be more than, less than, or equal to $165\, {\rm dB}$?

(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 $I_{20}$

\[\beta_{20}=10\,{\log \frac{I_{20}}{I_0}\ }\Rightarrow 82.5=10\,{\log \left(\frac{I_{20}}{I_0}\right)\ }\Rightarrow I_{20}={10}^{8.25}I_0\]

\[\therefore I_{20}=20\,I_1\Rightarrow I_1=\frac{{10}^{8.25}}{20}I_0\]

Now compute the intensity level for one violin as

\[\beta_1=10\,{\log \frac{I_1}{I_0}\ }\to \beta_1=10\,{\log \left(\frac{{10}^{8.25}}{20}\right)\ }=69.48\, {\rm dB}\]

(b) As previously, $I_{40}=40I_1$. Therefore,

\[I_{40}=10\,{\log \left(\frac{I_{40}}{I_0}\right)\ }=10\,{\log \left(\frac{40\frac{{10}^{8.25}}{20}I_0}{I_0}\right)\ \ }=85.5\ {\rm dB<165\ dB}\]

Your are trying to understand important instructions from someone who is soft-spoken. From your distance $r_1$ her voice sounds like an average whisper of $20\, {\rm dB}$. So you move to a position where you are a distance $r_2$ from her and the sound level is at $60\, {\rm dB}$.

(a) If the intensity is $I_1$ at distance $r_1$ and $I_2$ at distance $r_2$, calculate the ratio $I_2/I_1$.

(b) Use your answer to part (a) to calculate the ratio $r_2/r_1$.

An open organ pipe (open at both ends) has a harmonic with frequency of $440\, {\rm Hz}$. The next higher harmonic in the pipe has a frequency of $528\, {\rm Hz}$.

(a) Find the frequency of the fundamental.

(b) What is the length of the pipe?

A guitar string has a tension $100\, {\rm N\ }$ and has a mass of $7.0\, {\rm g}$ and is $0.45\, {\rm m}$ long.

(a) What is the speed of a wave on this string?

(b) What is the frequency of the third harmonic on this string?

(c) Two identical guitars, as described above, play a note in phase at their fundamental frequencies. If the guitars and a listener are positioned as shown in the drawing, what is the largest possible $d$, such that the listener hears no sound?

A train and a car are traveling in the same direction. The train's speed is $22.5\, {\rm m/s}$ and the car's speed is $9.5\, {\rm m/s}$. Assume the speed of the sound in air is $343\, {\rm m/s\ }$. if the train blows its whistle ($f=1500.0\, {\rm Hz}$), what is the frequency passengers in the car hear when:

(a) The train is approaching from behind the car.

(b) The train is moving away in front of the car.

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 is positive and all velocities in opposite direction is 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:

\[f_r=f_S\frac{v+u_r}{v+u_s}=1500\frac{343+9.5}{343+22.5}=1446.6\ {\rm Hz}\]

A carousel is $5.00$$\, {\rm m}$ in radius has a pair of $750\, {\rm Hz}$ sirens mounted on posts at opposite end of a diameter. The carousel rotates with an angular velocity of $0.800\, {\rm rad/s}$. A stationary listener is located at a distance from the carousel. The speed of the sound is $343\, {\rm m/s}$. What is the longest wavelength reaching the listener from the sirens?

A whistling train is moving with a velocity of $v_{train}= 25\, {\rm m/s}$ towards a station. The frequency of the train whistle is $500\,{\rm Hz}$.

a) Compute the frequency of the sound heard by a passenger on the platform in the station.

b) The sound of the whistling train is reflected by a nearby mountain and the reflected sound is heard by the train conductor. If the train moves towards the mountain with the same speed $v_{train}= 25\, {\rm m/s}$ as given above, what is the frequency of the reflected sound as heard by the train conductor?

c) What is the beat frequency between the emitted and reflected sound as heard by the train conductor?

Two eagles fly directly toward one another, the first at $15\, {\rm m/s\ }$and the second at $20\, {\rm m/s}$. Both birds screech, one emitting a frequency of $3200\, {\rm Hz}$ and the other at $3800\, {\rm Hz}$. What frequency do they receive if the speed of the sound is $330\, {\rm m/s}$ ?

\[f_{o1}=f_{s2}\frac{v\pm u_{o1}}{v\pm u_{s2}}=3800\frac{330+15}{330-20}=4229\ {\rm Hz}\]

In the case two, the eagle one is source and eagle two is receiver or object

\[f_{o2}=f_{s1}\frac{v\pm u_{o2}}{v\pm u_{s1}}=3200\frac{330+20}{330-15}=3556\ {\rm Hz\ }\]

An ambulance is traveling with its siren on. Initially you are standing still and observe a frequency of $0.9$ times the frequency as compared to what the driver of the ambulance hears. $v_{sound}=343\, {\rm m/s\ }$.

(a) Is the ambulance moving towards you or away from you?

(b) What is the speed of the ambulance?

(c) You later begin to run in the opposite direction as the ambulance at a speed of $3.0\, {\rm m/s\ }$. What frequency do you hear if the siren has a frequency of $900\, {\rm Hz}$?

The sound source of a ship's sonar system operates at a frequency of $22.0\,{\rm kHz}$. the speed of the sound in water (assumed to be at a uniform $20{\rm{}^\circ\!\,C}$) is $1482\, {\rm m/s}$.

a) What is the wavelength of the waves emitted by the source?

b) What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling directly toward the ship at $4.95\ {\rm m/s.}$ the ship is at rest in the water.

In a large tank, two tiny underwater speakers of radius $r=1.1\times {10}^{-3}\,{\rm m}$ are placed next to each other. The speakers vibrate at $1.2\, {\rm kHz}$ with the same amplitude (they are like vibrating beats). A microphone at the bottom of the tank records the superposition of sound from the two speakers. The mass of the each speakers is $0.015\,{\rm g}$ . the adiabatic bulk modulus of water is $B=2.2\times {10}^9\, {\rm N/m^2}$

a) What is the velocity of sound in the water?

b) What is the wavelength of the sound in the water?

c) One of the speakers is dropped and glides towards the bottom of the tank at constant velocity, while still emitting sound waves, what is the beat frequency recorded by the microphone? (you may model the speakers as spheres and assume the amplitude of the wave from each speakers is the same at the microphone)

At rest, a car's horn sounds at a frequency of $480\, {\rm Hz}$. The horn is sounded while the car is moving down the street. A bicyclist moving in the same direction with one-third the car's speed hears a frequency of $422\, {\rm Hz}$. What is the speed of the car? ( the speed of sound $v=345\, {\rm m/s}$)

When a certain string is clamped at both ends, the lowest four resonant frequencies are $50$,$100$,$150$, and $200 \mathrm {Hz}$. When the string is also clamped at its midpoint, the lowest four resonant frequencies are:

(a) 50, 100, 150, and 200 Hz

(b) 50, 150, 250, and 300 Hz

(c) 100, 200, 300, and 400 Hz

(d) 25, 50 75, and 100 Hz

(e) 75, 150, 225, and 300 Hz

The waves which produced in a string with both ends fixed is called standing waves. In this case, the length of 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 nth 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 travelling wave with velocity $v$ is $v=f\lambda $. Therefore, the frequency of nth mode is found as

\[f_n=n\frac{v}{2L}\ ,\ \ \left(n=1,2,3,\dots \right)\]

The frequency of $n=1$ case is called the fundamental frequency i.e. $f_1=v/2L$.

The $50\ \mathrm{Hz}$ frequency in above is fundamental frequency so using it we can find the ratio of $v/L$ as

\[f_1=\frac{v}{2L}\to 50=\frac{v}{2L}\to \frac{v}{L}=100\]

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.

A microphone of surface area $2.0\ {\mathrm{cm}}^{\mathrm{2}}$absorbs $1.1\ \mathrm{mW}$ of sound. What is the intensity of sound hitting the microphone?

(a) $2.2\times {10}^{-5}\mathrm{W/}{\mathrm{m}}^{\mathrm{2}}$

(b) $0.55\ \mathrm{W/}{\mathrm{m}}^{\mathrm{2}}$

(c) $2.2\ \mathrm{W/}{\mathrm{m}}^{\mathrm{2}}$

(d) $2.8\ \mathrm{W/}{\mathrm{m}}^{\mathrm{2}}$

(e) $5.5\ \mathrm{W/}{\mathrm{m}}^{\mathrm{2}}$

By definition of 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

\[I=\frac{1.1\times {10}^{-3}\mathrm{W}}{\mathrm{2.0\times }{\mathrm{10}}^{\mathrm{-}\mathrm{4}}{\mathrm{m}}^{\mathrm{2}}}=5.5\frac{\mathrm{W}}{{\mathrm{m}}^{\mathrm{2}}}\]

The correct answer is E.

Five organ pipes are described below. Which one has the highest fundamental frequency?

(a) A $2.3\, \mathrm m$ pipe with one end open and the other closed

(b) A $3.3\, \mathrm m$ pipe with one end open and the other closed

(c) A $1.6\, \mathrm m$ pipe with both ends open

(d) A $3.0\, \mathrm m$ pipe with both ends open

(e) A pipe in which the displacement nodes are $5\, \mathrm m$ apart

A particle with a charge of $5\times {10}^{-6}\mathrm{C}$ and a mass of $\mathrm{20\ g}$ moves uniformly with a speed of $\mathrm{7\ m/s}$ in a circular orbit around a stationary particle with a charge of $-5\times {10}^{-6}\mathrm{C}$. The radius of the orbit is:

(a) $0\ \mathrm{m}$

(b) $0.23\ \mathrm{m}$

(c) $0.62\ \mathrm{m}$

(d) $1.6\ \mathrm{m}$

(e) $4.4\ \mathrm{m}$

In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance $D$ must be changed to:

(a) $D/2$

(b) $D/\sqrt{2}$

(c) $D\sqrt{2}$

(d) $2D$

(e) $4D$

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

\[\mathrm{\Delta }y=y_{m+1}-y_m=D\frac{\left(m+1\right)\lambda }{d}-D\frac{m\lambda }{d}=D\frac{\lambda }{d}\]

Since the source does not change $\lambda ={\lambda }^{'}$ so

\[\frac{\mathrm{\Delta }y}{\mathrm{\Delta }y^{'}}=\frac{D}{D^{'}}\frac{d^{'}}{d}\Rightarrow 1=\frac{D}{D^{'}}\frac{2d}{d}\Rightarrow D^{'}=2D\]

The correct answer is D.

Category : Waves

**Most useful formula in Waves:**

Sound wave in an ideal gas: $$v=\sqrt{\frac{\gamma RT}{M}}$$

Open pipe: $f_n=\frac{nv}{2L}$

Closed pipe: $f_n=\frac{nv}{4L}$

Doppler effect:

\[ f_O=f_s\frac{v_O \pm u_O}{v_s \pm u_s}\]

Sound intensity Level: $$\beta=(10\, \rm{dB})\, \rm{\log}\,\frac{I}{1.0 \times 10^{-12} \rm{\frac{W}{m^2}}}$$

The intensity of the light transmitted through the analyzer is (Malus's law):

\[I=I_{max}\cos^2 \phi\]

where $I_{max}$ is the maximum intensity at $\phi=0$ and $I$ is the amount of transmitted light at angle $\phi$.

Number Of Questions : 37

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