# Sound Waves

## Concept Of Physics

### H C Verma

1   A steel tube of length $1.00\ m$ is struck at one end. A person with his ear close to the other end hears the sound of the blow twice, one travelling through the body of the tube and the other through the air in the tube. Find the time gap between the two hearings. Use the table in the text for speeds of sound in various substances.

##### Solution :

So, $t = t_{1}\ -\ t_{2} = \big(\frac{1}{330}\ -\ \frac{1}{5200}\big) = 2.75 \times 10^{-1}\ sec = 2.75\ ms$

$V_{air} = 230\ m/s. V_{5} = 5200\ m/s$. Here $s = 7\ m$

2   At a prayer meeting, the disciples sing $JAI-RAM\ JAI-RAM$. The sound amplified by a loudspeaker comes back after reflection from a building at a distance of $80\ m$ from the meeting. What maximum time interval can be kept between one $JAI-RAM$ and the next $JAI-RAM$ so that the echo does not disturb a listener sitting in the meeting. Speed of sound in air is $320\ m/s$.

##### Solution :

$v = 320 m/s$

Here given $S = 80 m \times 2 = 160\ m$

So minimum time interval will be $\\$ $t = \frac{5}{v} = \frac{160}{320} = 0.5$ seconds.

3   A man stands before a large wall at a distance of $50\cdot0\ m$ and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap $10$ times during every $3$ seconds, find the velocity of sound in air.

##### Solution :

He ha to clap $10$ times in $3$ seconds. $\\$ So time interval between two clap = ($\frac{3}{10}$ second). $\\$ So the taken go the wall = $\big(\frac{3}{2} \times 10\big)$ = $\frac{3}{20}$ seconds. $\\$ = $333\ m/s$.

4   A person can hear sound waves in the frequency range $20\ Hz$ to $20\ kHz$. Find the minimum and the maximum wavelengths of sound that is audible to the person. The speed of sound is $360\ m/s$.

##### Solution :

b) For minimum wavelength, $n = 20 KHz$ $\\$ $\therefore \lambda = \frac{360}{(20 \times 10^3)} = 18 \times 10^{-3}\ m = 18\ mm$ $\\$ $\Rightarrow x = (\frac{v}{n}) = \frac{360}{20} = 18\ m$

a) For minimum wavelength $n = 20\ KHz$ $\\$ as $\big(\eta\ \infty\ \frac{1}{\lambda}\big)$

5   Find the minimum and maximum wavelengths of sound in water that is in the audible range $(20\ -\ 20000\ Hz)$ for an average human ear. Speed of sound in water $= 1450\ m/s$.

##### Solution :

b) for minimum wavelength $n$ should be minimum $\\$ $\Rightarrow v = n\lambda \Rightarrow \lambda = \frac{v}{n} \Rightarrow \frac{1450}{20} = 72.5\ m$

a) For minimum wavelength $n = 20\ KHz$ $\\$ $\Rightarrow v = n\lambda \Rightarrow \lambda = \big(\frac{1450}{20 \times 10^3}\big) = 7.25\ cm$

6   Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker. (a) Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is $20\ cm$. (b) Sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one tenth of the diameter of the speaker described above. Take the speed of sound to be $340\ m/s$.

##### Solution :

$N = \frac{v}{\lambda} \Rightarrow \frac{340}{2 \times 10^{-2}} = 17.00\ Hz = 17\ KH_{2}$ (Because $\lambda = 2 cm = 2 \times 10_{-2}\ m$)

According to the question, $\\$ a) $\lambda = 20\ cm \times 10 = 200\ cm = 2\ m$ $\\$ $v = 340\ m/s$ $\\$ So, $n = \frac{340}{2} = 170\ Hz$.

7   Ultrasonic waves of frequency $4.5\ MHz$ are used to detect tumour in soft tissues. The speed of sound in tissue is $1\cdot5\ km/s$ and that in air is $340\ m/s$. Find the wavelength of this ultrasonic wave in air and in tissue.

##### Solution :

b) $V_{tissue} = 1500\ m/s \Rightarrow \lambda = \big(\frac{1500}{4.5}\big)\times 10^{-6} = 3.3 \times 10^{-4}\ m$.

a) Given $V_{air} = 340\ m/s, n = 4.5 \times 10^{6}\ Hz$ $\\$ $\Rightarrow \lambda_{air} = \big(\frac{340}{4.5}\big) \times 10^{-6} = 7.36 \times 10^{-5}\ m$.

8   The equation of a travelling sound wave is $y = 6\cdot0\ sin\ (600\ t - 1\cdot8 x)$ where $y$ is measured in $10^{-5}\ m$, $t$ in second and $x$ in metre. (a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. (b) Find the ratio of the velocity amplitude of the particles to the wave speed.

##### Solution :

Again, $\lambda = \frac{2\pi}{1.8}$ and $T = \frac{2\pi}{600} \Rightarrow$ wave speed $= v = \frac{\lambda}{T} = \frac{600}{1.8} = \frac{1000}{3}\ m/s$

So, $\frac{r_{y}}{\lambda} = \frac {6.0 \times (1.8) \times 10^{-5}\ m/s}{2\pi} = 1.7 \times 10^{-5}\ m/s$

Here $V_{y} = 3600 \times 10^{-5}\ m/s$

Here given $r_{y} = 6.0 \times 10^{-5}\ m$ $\\$ a) Given $\frac{2\pi}{\lambda} = 1.8 \Rightarrow \lambda = (\frac{2\pi}{1.8})$

So the ratio of $\big(\frac{V_{y}}{v}\big) = \frac{3600 \times 3 \times 10^{-5}}{1000}$.

b) Let, velocity amplitude = $V_{y}$ $\\$ $V = \frac{dy}{dt} = 3600\ cos\ (600 t - 1.8) \times 10^{-5}\ m/s$

9   A sound wave of frequency $100\ Hz$ is travelling in air. The speed of sound in air is $350\ m/s$. (a) By how much is the phase changed at a given point in $2\cdot5\ ms$ ? (b) What is the phase difference at a given instant between two points separated by a distance of $10\cdot0\ cm$ along the direction of propagation ?

##### Solution :

b) In the second case, Given $\Delta{\eta} = 10\ cm = 10^{–1}\ m$ $\\$ So, $\phi = \dfrac{2\pi}{x} \Delta{x} \dfrac{2\pi \times 10{-1}}{(350 /100)} = \dfrac{2\pi}{35}$

Here given $n = 100,\ v = 350\ m/s.$ $\\$ $\Rightarrow = \dfrac{v}{n} = \dfrac{350}{100} = 3.5\ m$ $\\$ In $2.5\ ms$, the distance travelled by the particle is given by, $\\$ $\Delta{x} = 350 \times 2.5 \times 10^{–3}$ $\\$ So, phase difference $\phi = \dfrac{2\pi}{\lambda} \times \Delta{x} \Rightarrow \dfrac{2\pi}{350/100} \times 350 \times 2.5 \times 10^{-3} = \big(\dfrac{\pi}{2}\big)$

10   Two point sources of sound are kept at a separation of $10\ cm$. They vibrate in phase to produce waves of wavelength $5\cdot0\ cm$. What would be the phase difference between the two waves arriving at a point $20\ cm$ from one source (a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources ?

##### Solution :

a) Given $\Delta{x} = 10\ cm,\ \lambda = 5.0\ cm$ $\\$ $\Rightarrow \phi = \dfrac{2\pi}{\lambda} \times \Delta{\eta}={2\pi}{5} \times 10 = 4\pi$ $\\$ So phase difference is zero. $\\$ b) Zero, as the particle is in same phase because of having same path.

11   Calculate the speed of sound in oxygen from the following data. The mass of $22\cdot4\ litre$ of oxygen at $STP$ $(T = 273\ K\ and\ p = 1\cdot0 \times 10^5 N/m^2)$ is $32\ g$, the molar heat capacity of oxygen at constant volume is $C_{v} = 2\cdot5\ R$ and that at constant pressure is $C_{p} = 3\cdot5\ R$.

##### Solution :

$V = 22.4 litre = 22.4 \times 10^{–3} m^3$ $\\$ $\dfrac{C}{C_v} = r = \dfrac{3.5 R}{2.5 R} = 1.4$ $\\$ $\Rightarrow V = \sqrt{\dfrac{rp}{f}} = \sqrt{\dfrac{1.2 \times 1.0 \times 10^{-5}}{32/22.4}} = 310 m/s$ $\ \ \ \ \ \ \$ (because $\rho = m/v$)

Given that $p = 1.0 \times 10{5} N/m^2,\ T = 273 K,\ M = 32 g = 32 \times 10^{–3} kg$ $\\$

12   The speed of sound as measured by a student in the laboratory on a winter day is $340\ m/s$ when the room temperature is $17°C$. What speed will be measured by another student repeating the experiment on a day when the room temperature is $32°C$ ?

##### Solution :

$T_1 = 273 + 17 = 290 K,\ T_2 = 272 + 32 = 305 K$

$= 340 \times \sqrt{\dfrac{305}{290}} = 349\ m/s$

$V_1 = 330\ m/s, V2 = ?$

We know $v\ \infty\ \sqrt{T}$ $\\$ $\dfrac{\sqrt{V_1}}{\sqrt{V_2}} = \dfrac{\sqrt{T_1}}{\sqrt{T_2}} \Rightarrow V_2 = \dfrac{V_1 \times \sqrt{T_2}}{\sqrt{T_1}}$

13   At what temperature will the speed of sound be double of its value at $0°C$ ?

##### Solution :

We know that $V\ \infty\ \sqrt{T} \Rightarrow \dfrac{T_2}{T_1} = \dfrac{V_{2}^{2}}{V_{1}^{2}} \Rightarrow T_2 = 273 \times 2^2 = 4 \times 273 K$

$T_1 = 273 \ \ \ \ \ \ V_2 = 2V_1$ $\\$ $V_1 = v \ \ \ \ \ \ \ T2 =$?

So temperature will be $(4 \times 273)\ –\ 273 = 819°c$.

14   The absolute temperature of air in a region linearly increases from $T_1$ to $T_2$ in a space of width $d$. Find the time taken by a sound wave to go through the region in terms of $T_1,$ $T_vd$ and the speed $v$ of sound at $273\ K$. Evaluate this time for $T_1 = 280\ K,\ T_2 = 310\ K,\ d = 33\ m$ and $v = 330\ m/s$.

##### Solution :

We know that $V\ \infty\ \sqrt{T} \Rightarrow \dfrac{V_t}{V} = \sqrt{\dfrac{T}{273}} \Rightarrow VT = v\sqrt{\dfrac{T}{273}}$ $\\$ $\Rightarrow dt= \dfrac{dx}{V_T} = dfrac{du}{V} \times\sqrt{\dfrac{273}{T}}$ $\\$ $\Rightarrow t = \dfrac{273}{V} \int_{0}^{d}\dfrac{dx}{[T_1 + (T_2 - T_1/dx)]\dfrac{1}{2}}$ $\\$ $= \dfrac{\sqrt{273}}{V} \times \dfrac{2d}{T_2 - T_1} [T_1 + \dfrac{T_2 - T_1}{d}x]_{0}^{d} = \big(\dfrac{2d}{V}\big)\big(\dfrac{\sqrt{273}}{T_2 - T_1}\big) \times \sqrt{T_2} \sqrt{T_1}$ $\\$ $= T = \dfrac{2d}{V} \dfrac{\sqrt{273}}{\sqrt{T_2} + \sqrt{T_1}}$ $\\$ Putting the given value we get $\\$ $= \dfrac{2 \times 33}{330} \dfrac{\sqrt{273}}{\sqrt{280} + \sqrt{310}} = 96\ ms$

The variation of temperature is given by, $\\$ $T = T_1 + \dfrac{(T2 - T2 )}{d}x \ \ \ \ \ \ \ \ \ \ \ \ …(1)$

15   Find the change in the volume of PO litre kerosene when it is subjected to an extra pressure of $2.0 \times 10^{5}\ N/m^2$ from the following data. Density of kerosene = 800 kg/m 3 and speed of sound in kerosene = $1330\ m/s$.

##### Solution :

We know that $v = \dfrac{\sqrt{K}}{\rho}$ $\\$ Where $K$ = bulk modulus of elasticity $\\$ $\Rightarrow K = v^2\ \rho = (1330)^2 \times 800\ N/m^2$ $\\$ We know $K = \big(\dfrac{F/A}{\Delta{V}/V}\big)$ $\\$ $\Rightarrow \Delta{V} = \dfrac{Pressures}{K} = \dfrac{2 \times 10^{5}}{1330 \times 1330\times 800}$ $\\$ So, $\Delta{V} = 0.15\ cm^{3}$

16   Calculate the bulk modulus of air from the following data about a sound wave of wavelength $35\ cm$ travelling in air. The pressure at a point varies between $(1\cdot0 \times 10^{5} ± 14)\ Pa$ and the particles of the air vibrate in simple harmonic motion of amplitude $5\cdot5 \times 10^{-6}\ m$.

##### Solution :

We know that, $\\$ Bulk modulus $B = \dfrac{\Delta{p}}{(\Delta{V}/V)} = \dfrac{P_0\lambda}{2\pi{S_0}}$ $\\$ Where $P_0$ = pressure amplitude $\Rightarrow P_0 = 1.0 \times 10^{5}$ $\\$ $S_0$ = displacement amplitude $\Rightarrow S_0 = 5.5 \times 10^{–6}\ m$ $\\$ $\Rightarrow \dfrac{14 \times 35 \times 10^{-2}\ m}{2\pi(5.5) \times 10^{-6}\ m} = 1.4 \times 10^{5} N/m^2$

17   A source of sound operates at $2\cdot0\ kHz,\ 20\ W$ emitting sound uniformly in all directions. The speed of sound in air is $340\ m/s$ and the density of air is $1\cdot2\ kg/m^3$. (a) What is the intensity at a distance of $6\cdot0\ m$ from the source ? (b) What will be the pressure amplitude at this point ? (c) What will be the displacement amplitude at this point ?

##### Solution :

a) Here given $V_{air} = 340\ m/s.,\ Power = \dfrac{E}{t} = 20\ W$ $\\$ $f = 2,000 Hz,\ \rho = 1.2 kg/m^3$ $\\$ So, intensity $I = \dfrac{E}{t.A}$ $\\$ $= \dfrac{20}{3\pi{r^2}} = \dfrac{20}{4 \times \pi \times 6^2} = 44\ mw/m^2$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (because $r = 6\ m$) $\\$ b) We know that $I = \dfrac{p_{0}^{2}}{2\rho V_{air}} \Rightarrow P_0 = \sqrt{1 \times 2\rho V_{air}}$ $\\$ $= \sqrt{2 \times 1.2 \times 340 \times 44 \times 10{-3}} = 6.0 N/m^2$ $\\$ c) We know that $I = 2\pi^2S_{0}^{2}v^2\rho V$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \$ where $S_0$ = displacement amplitude $\\$ $\Rightarrow S_0 = \sqrt{\dfrac{I}{\pi^2 \rho^2 \rho V_{air}}}$ $\\$ Putting the value we get $S_g = 1.2 \times 10^{–6}\ m$.

18   The intensity of sound from a point source is $1\cdot0 \times 10^{-6}\ W/m^2$ at a distance of $5\cdot0\ m$ from the source. What will be the intensity at a distance of $25\ m$ from the source ?

##### Solution :

Here $I_1 = 1.0 \times 10^{–8}\ W_1/m^2 ;\ I_2 = ?$ $\\$ $r_1 = 5.0 m,\ r_2 = 25\ m.$ $\\$ We know that $I\ \infty\ \dfrac{1}{r^2}$ $\\$ $\Rightarrow I_1r_{1}^{2} = I_2 r_{2}^{2} \Rightarrow \dfrac{ I_1r_{1}^{2}}{r_{2}^{2}}$ $\\$ $= \dfrac{1.0 \times 10^{-8} \times 25}{625} = 4.0 \times 10^{-10}\ W/m^2$

19   The sound level at a point $5.0\ m$ away from a point source is $40\ dB$. What will be the level at a point $50\ m$ away from the source ?

##### Solution :

$\Rightarrow \dfrac{\beta_{A} - \beta_{B}}{10} = 2 \Rightarrow \beta_{A} - \beta_{B} = 10^2$

$\beta_{A} = 10\ log \dfrac{I_A}{I_0},\ \beta_{B} = 10\ log \dfrac{I_A}{I_0}$

$\dfrac{I_{A}}{I_{B}} = \dfrac{r_{B}^{2}}{r_{A}^{2}} = \big(\dfrac{50}{5}\big)^2 \Rightarrow 10^{(\beta_{A}\beta_{B})} = 10^2$

We know that $\beta = 10\ log_{10} \big(\dfrac{I}{I_0}\big)$

$\Rightarrow \beta_{B} = 40 -20 = 20\ d\beta$

$\dfrac{I_{A}}{I_{0}} = 10^{(\beta_{A}/ 10)} \Rightarrow \dfrac{I_{B}}{I_{o}} = 10^{(\beta_{B}/ 10)}$

20   If the intensity of sound is doubled, by how many decibels does the sound level increase ?

##### Solution :

We know that, $\beta = 10\ log_{10} \dfrac{J}{I_{0}}$ $\\$ According to the questions, $\\$ $\beta_{A} = 10\ log_{10}\big(\dfrac{2I}{I_0}\big)$ $\\$ $\Rightarrow \beta_{B} - \beta_{A} = 10\ log \Big(\dfrac{2I}{I}\Big) = 10 \times 0.3010 = 3\ dB$

21   Sound with intensity larger than $120\ dB$ appears painful to a person. A small speaker delivers $2\cdot0\ W$ of audio output. How close can the person get to the speaker without hurting his ears ?

##### Solution :

If sound level $= 120\ dB,$ then $I =$ intensity $= 1\ W/m^2$ $\\$ Given that, audio output $= 2W$ $\\$ Let the closest distance be $x$. $\\$ So, intensity $= \Big(\dfrac{2}{4\pi{x^2}}\Big) = 1 \Rightarrow x^2 = \Big(\dfrac{2}{2\pi}\Big) \Rightarrow x = 0.4\ m = 40\ cm$ $\\$

22   If the sound level in a room is increased from $50\ dB$ to $60\ dB$, by what factor is the pressure amplitude increased ?

##### Solution :

$\beta_{1} = 50\ dB,\ \beta_{2} = 60\ dB$ $\\$ $\therefore I_1 = 10^{–7} W/m^2,\ I_2 = 10^{–6}\ W/m^2$ $\\$ (because $\beta = 10\ log_10 \Big(\dfrac{I}{I_0})$, where $I_0 = 10^{–12} W/m^2$) $\\$ Again, $\dfrac{I_2}{I_1} = \Big(\dfrac{p_2}{p_1}\Big)^2 = \Big(\dfrac{10^{–6}}{10^{–7}}\Big) = 10$ (where $p =$ pressure amplitude) $\\$ $\therefore \Big(\dfrac{p_2}{p_1}\Big) = \sqrt{10}.$

23   The noise level in a class-room in absence of the teacher is $50\ dB$ when $50$ students are present. Assuming that on the average each student outputs same sound energy per second, what will be the noise level if the number of students is increased to $100$ ?

##### Solution :

$\beta_{B}\ – \beta_{A} = 10 log_{10} \dfrac{50I}{I_{0}} - 10log_{10}\Big(\dfrac{100 I}{I_0}\Big)$

So, $\beta_{A} = 50 + 3 = 53 dB.$

Let the intensity of each student be $I$. $\\$ According to the question $\\$ $\beta_{A} = 10\ log_{10}\dfrac{50 I}{I_0} ; \beta_{B} = 10log_{10}\Big(\dfrac{100 I}{I_0}\Big)$

$= 10 log \Big(\dfrac{100I}{50I}\Big) = 10 log_{10}2 = 3$

24   In a Quincke's experiment the sound detected is changed from a maximum to a minimum when the sliding tube is moved through a distance of $2\cdot50\ cm.$ Find the frequency of sound if the speed of sound in air is $340\ m/s$.

##### Solution :

Distance between tow maximum to a minimum is given by, $\lambda/4 = 2.50\ cm$ $\\$ $\Rightarrow \lambda = 10 cm = 10^{-1}\ m$ $\\$ We know, $V = nx$ $\\$ $\Rightarrow = \dfrac{V}{\lambda} = \dfrac{340}{10^{-1}} = 3400\ Hz = 3.4\ kHz$

25   In a Quincke's experiment, the sound intensity has a minimum value I at a particular position. As the sliding tube is pulled out by a distance of $16\cdot5\ mm$, the intensity increases to a maximum of $9I$. Take the speed of sound in air to be $330\ m/s$. (a) Find the frequency of the soundsource. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.

##### Solution :

So, $\dfrac{A_1 + A_2}{A_1 + A_2} = \dfrac{3}{4} \Rightarrow \dfrac{A_1}{A_2} = \dfrac{2}{1}$

a) According to the data $\lambda/4 = 16.5\ mm \Rightarrow \lambda = 66 mm = 66 \times 10^{-6 = 3}\ m.$ $\\$ $\Rightarrow n = \dfrac{V}{\lambda} = \dfrac{330}{66 \times 10{-3}} = 5 kHz$ $\\$ b) $I_{minimum} = K(A_1 – A_2)^2 = I \Rightarrow A_1 – A_2 = 11$ $\\$ $I_{maximum} = K(A_1 + A_2)^2 = 9 \Rightarrow A_1 + A_2 = 31$

So, the ratio amplitudes is $2$.

26   Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place $6\cdot0\ m$ from one of the speakers and $6\cdot4\ m$ from the other. If the sound signal is continuously varied from $500\ Hz$ to $5000\ Hz$, what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air = $320\ m/s$.

##### Solution :

The path difference of the two sound waves is given by, $\\$ $\Delta{L} = 6.4 – 6.0 = 0.4\ m$ $\\$ The wavelength of either wave $= \lambda = \dfrac{V}{\rho} = \dfrac{320}{\rho} (m/s)$ $\\$ For destructive interference $\Delta{L} = \dfrac{(2n + 1)\lambda}{2}$ where n is an integers. $\\$ or $0.4\ m = \dfrac{2n + 1}{2} \times \dfrac{320}{\rho}$ $\\$ $\Rightarrow \rho = n = \dfrac{}{} = 8000\dfrac{}{}Hz = (2n + 1) 400\ Hz$ $\\$ Thus the frequency within the specified range which cause destructive interference are $1200\ Hz,\ 2000\ Hz,\ 2800\ Hz,\ 3600\ Hz$ and $4400\ Hz$.

27   A source of sound $S$ and a detector $D$ are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line $SD$ as shown in figure (16-E1). It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of $20\ cm$. Find the frequency of the sound emitted. Velocity of sound in air is $336\ m/s$.

##### Solution :

According to the given data $\\$ $V = 336\ m/s$, $\\$ $\lambda/4 =$ distance between maximum and minimum intensity. $\\$ $= (20cm) \Rightarrow = 80\ cm$ $\\$ $\Rightarrow n =$ frequency $= \dfrac{V}{\lambda} = \dfrac{336}{80 \times 10{-2}} = 420\ Hz$

28   A source $S$ and a detector $D$ are placed at a distance $d$ apart. A big cardboard is placed at a distance $\sqrt{2d}$ from the source and the detector as shown in figure $(16-E2)$. The source emits a wave of wavelength = $d/2$ which is received by the detector after reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave ?

##### Solution :

$= 2\sqrt{\dfrac{}{}} + (\sqrt{2d} + x)^2 - d = \dfrac{d}{4}\Big(2d + \dfrac{d}{4}\Big)$ $\\$ $\rightarrow \Big(\dfrac{d}{2}\Big)^2 = \dfrac{169d^2}{64} \Rightarrow = \dfrac{153}{64}d^2$ $\\$ $\Rightarrow \sqrt{2d} + x = 1.54\ d \Rightarrow x = 1.54 d - 1.414 d = 0.13 d.$

Here given $\lambda = \dfrac{d}{2}$ $\\$ Initial path difference is given by $= 2\sqrt{\Big(\dfrac{d}{2}\Big) + 2d^2} - d$ $\\$ If it is now shifted a distance $x$ then path difference will be

29   Two stereo speakers are separated by a distance of $2\cdot40\ m$. A person stands at a distance of $3\cdot20\ m$ directly in front of one of the speakers as shown in figure $(16-E3)$. Find the frequencies in the audible range $(20 - 2000 Hz)$ for which the listener will hear a minimum sound intensity. Speed of sound in air = $320\ m/s$.

##### Solution :

As shown in the figure the path differences $2.4 = \Delta{x} = \sqrt{(3.2)^2 + (2.4)^2} - 3.2$ $\\$ Again, the wavelength of the either sound waves = $\dfrac{320}{\rho}$ $\\$ We know, destructive interference will be occur $\\$ If $\Delta{x} = \dfrac{(2n + 1)\lambda}{2}$ $\\$ $\Rightarrow \sqrt{(3.2)^2 + (2.4)^2 - (3.2)} = \dfrac{(2n + 1)}{2}\dfrac{320}{\rho}$ $\\$ Solving we get, $\\$ $\Rightarrow V = \dfrac{(2n + 1)400}{2} = 200(2n + 1)$ $\\$ where $n = 1, 2, 3, .......... 49$. (audile region)

30   Two sources of sound, $S1$ and $S2$ , emitting waves of equal wavelength $20\cdot0\ cm$, are placed with a separation of $20\cdot0\ cm$ between them. A detector can be moved on a line parallel to $S_{1}$ $S_{2}$ and at a distance of $20\cdot0\ cm$ from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance through which the detector should be shifted to detect a minimum of sound.

##### Solution :

According to the data, $\\$ $\lambda = 20\ cm,\ S_1S_2 = 20\ cm,\ BD = 20\ cm$ $\\$ Let the detector is shifted to left for a distance $x$ for hearing the minimum sound. So path difference $AI = BC – AB$ $\\$ $= \sqrt{(20)^2 + (10 + x)^2} - \sqrt{(20)^2 + (10 - x)^2}$ $\\$ So the minimum distances hearing for minimum $\\$ $= \dfrac{(2n + 1)\lambda}{2} = \dfrac{\lambda}{2} = \dfrac{20}{2} = 10\ cm$ $\\$ $\Rightarrow \sqrt{(20)^2 + (10 + x)^2} - \sqrt{(20)^2 + (10 - x)^2} = 10$ solving we get $x = 12.0\ cm$

31   Two speakers $S_{1}$ and $S_{2}$ , driven by the same amplifier, are placed at $y$ = $1\cdot0\ m$ and $y$ = $-1\cdot0\ m$ $(figure 16-E4)$. The speakers vibrate in phase at $600\ Hz$. A man stands at a point on the $X$-axis at a very large distance from the origin and starts moving parallel to the $Y$-axis. The speed of sound in air is $330\ m/s$. (a) At what angle $\theta$ will the intensity of sound drop to a minimum for the first time ? (b) At what angle will he hear a maximum of sound intensity for the first time ? (c) If he continues to walk along the line, how many more maxima can he hear ?

##### Solution :

c) For more maxima, $\\$ $\dfrac{yd}{D} = 2\lambda,\ 3\lambda,\ 4\lambda, ..…$ $\\$ $\Rightarrow \dfrac{y}{D} = \theta = 32°,\ 64°,\ 128°$ $\\$ But since, the maximum value of $\theta$ can be $90°$, he will hear two more maximum i.e. at $32°$ and $64°.$

Let $OP = D,\ PQ = y \Rightarrow \theta = \dfrac{y}{R} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ …(1)$ $\\$ Now path difference is given by, $x = S_2Q – S_1Q = \dfrac{yd}{D}$ $\\$ Where $d = 2\ m$ $\\$ [The proof of $x = \dfrac{yd}{D}$ is discussed in interference of light waves]

b) For minimum intensity, $x = 2n(\dfrac{\lambda}{2}\Big)$ $\\$ $\dfrac{yd}{D} = \lambda \Rightarrow \dfrac{y}{D} = \theta = \dfrac{\lambda}{D} = \dfrac{0.55}{2} = 0.275$ rad $\\$ $\therefore \theta = 16°$

Given, $F = 600\ Hz,$ and $v = 330\ m/s \Rightarrow \dfrac{v}{f} = \dfrac{330}{600} = 0.55\ mm$

a) For minimum intensity, $x = (2n + 1)(\lambda/2)$ $\\$ $\dfrac{yd}{D} = \dfrac{\lambda}{2}$ [for minimum $y,\ x = \dfrac{\lambda}{2}$] $\\$ $\therefore \dfrac{y}{D} = \theta = \dfrac{\lambda}{2} = \dfrac{0.55}{4} = 0.1375$ rad $= 0.1375 \times (57.1)° = 7.9°$

32   Three sources of sound $S_{1}\ S_{2}$ and $S_{3}$ of equal intensity are placed in a straight line with $S_{1}S_{2}$ = $S_{2}S_{3}$, $(figure\ 16-E5)$. At a point $P$, far away from the sources, the wave coming from $S_{2}$ is $120°$ ahead in phase of that from $S_{1}$. Also, the wave coming from $S_{3}$ is $120°$ ahead of that from $S_{2}$. What would be the resultant intensity of sound at $P$?

##### Solution :

Because the $3$ sources have equal intensity, amplitude are equal $\\$ So, $A_1 = A_2 = A_3$ $\\$ As shown in the figure, amplitude of the resultant $= 0$ (vector method) $\\$ So, the resultant, intensity at $B$ is zero.

33   Two coherent narrow slits emitting sound of wavelength $\lambda$ in the same phase are placed parallel to each other at a small separation of $2\lambda$. The sound is detected by moving a detector on the screen $\sum$ at a distance $D(\gg\lambda)$ from the slit $S_{1}$ as shown in $figure\ (16-E6)$. Find the distance $x$ such that the intensity at $P$ is equal to the intensity at $O$.

##### Solution :

The two sources of sound $S_1$ and $S_2$ vibrate at same phase and frequency. $\\$ Resultant intensity at $P = I_0$ $\\$ a) Let the amplitude of the waves at $S_1$ and $S_2$ be $‘r’$. $\\$ When $\theta = 45°$, path difference $= S_1P – S_2P = 0$ (because $S_1P = S_2P$) $\\$ So, when source is switched off, intensity of sound at $P$ is $\dfrac{I_0}{4}$ $\\$. b) When $\theta = 60°$, path difference is also $0.$ Similarly it can be proved that, the intensity at $P$ is $\dfrac{I_0}{4}$ when one is switched off.

34   $Figure (16-E7)$ shows two coherent sources $S_{1}$ and $S_{2}$ which emit sound of wavelength $\lambda$ in phase. The separation between the sources is $3\lambda$. A circular wire of large radius is placed in such a way that $S_{1}S_{2}$ lies in its plane and the middle point of $S_{1}S_{2}$ is at the centre of the wire. Find the angular positions $\theta$ on the wire for which constructive interference takes place.

##### Solution :

If $V = 340\ m/s,\ I = 20\ cm = 20 \times 10^{–2}\ m$ $\\$ Fundamental frequency $= \dfrac{V}{21} = \dfrac{340}{2 \times 20 \times 10^{-2}} = 850\ Hz$ $\\$ We know first over tone $= \dfrac{2V}{21} = \dfrac{2 \times 340}{2 \times 20 \times 10^{-2}}$ (for open pipe) = $1750\ Hz$ $\\$ Second over tone $= 3 \Big(\dfrac{V}{21}\Big) = 3 \times 850 = 2500\ Hz$

35   Two sources of sound $S_{1}$ and $S_{2}$ vibrate at same frequency and are in phase $(figure 16-E8)$. The intensity of sound detected at a point P as shown in the figure is $I_{0}$. (a) If $\theta$ equals $45°$, what will be the intensity of sound detected at this point if one of the sources is switched off ? (b) What will be the answer of the previous part if $\theta$ = $60°$ ?

##### Solution :

According to the questions $V = 340\ m/s,\ n = 500\ Hz$ $\\$ We know that $\dfrac{V}{4I}$ $\ \ \ \ \$ (for closed pipe) $\\$ $\Rightarrow I = \dfrac{340}{4 \times 500}\ m = 17\ cm$

36   Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length $20\ cm$. Speed of sound in air is $340\ m/s$.

##### Solution :

Here given distance between two nodes is $= 4. 0\ cm$, $\\$ $\Rightarrow \lambda = 2 \times 4.0 = 8\ cm$ $\\$ We know that $v = n\lambda$ $\\$ $\Rightarrow \eta = \dfrac{328}{8 \times 10{-2}} = 4.1\ Hz$

37   A closed organ pipe can vibrate at a minimum frequency of $500\ Hz$. Find the length of the tube. Speed of sound in air = $340\ m/s$.

##### Solution :

$V = 340\ m/s$ $\\$ Distances between two nodes or antinodes $\\$ $\Rightarrow \dfrac{\lambda}{4} = 25\ cm$ $\\$ $\Rightarrow \lambda = 100\ cm = 1\ m$ $\\$ $\Rightarrow n = \dfrac{v}{\lambda} = 340\ Hz$.

38   In a standing wave pattern in a vibrating air column, nodes are formed at a distance of $4\cdot0\ cm$. If the speed of sound in air is $328\ m/s$, what is the frequency of $73$ the source ?

##### Solution :

Here given that $1 = 50\ cm,\ v = 340\ m/s$ $\\$ As it is an open organ pipe, the fundamental frequency $f_1 = (\dfrac{v}{21})$ $\\$ $= \dfrac{340}{2 \times 50 \times 10{-2}} = 340\ Hz$ $\\$ So, the harmonies are $\\$ $f_3 = 3 \times 340 = 1020 Hz$ $\\$ $f_5 = 5 \times 340 = 1700,\ f_6 = 6 \times 340 = 2040\ Hz$ $\\$ so, the possible frequencies are between $1000\ Hz$ and $2000\ Hz$ are $1020,\ 1360,\ 1700$.

39   The separation between a node and the next antinode in a vibrating air column is $25\ cm$. If the speed of sound in air is $340\ m/s$, find the frequency of vibration of the air column.

##### Solution :

Here given $I_2 = 0.67\ m,\ l_1 = 0.2\ m, f = 400\ Hz$ $\\$ We know that $\\$ $\lambda = 2(l_2 – l_1) \Rightarrow \lambda = 2(62 – 20) = 84\ cm = 0.84\ m$. $\\$ So, $v = n\lambda = 0.84 \times 400 = 336\ m/s$ $\\$ We know from above that, $\\$ $l_1 + d = \dfrac{\lambda}{4} \Rightarrow d = \dfrac{\lambda}{4} – l_1 = 21 – 20 = 1\ cm.$

40   A cylindrical metal tube has a length of $50\ cm$ and is open at both ends. Find the frequencies between $1000\ Hz$ and $2000\ Hz$ at which the air column in the tube can resonate. Speed of sound in air is $340\ m/s$.

##### Solution :

According to the questions $\\$ $f_1$ first overtone of a closed organ pipe $P_1 = \dfrac{3v}{4l} = \dfrac{3 \times V}{4 \times 30}$ $\\$ $f_2$ fundamental frequency of a open organ pipe $P_2 = \dfrac{V}{2I_2}$ $\\$ Here given $\dfrac{3V}{4 \times 30} = \dfrac{V}{2I_2} \Rightarrow I_2 = 20\ cm$ $\\$ $\therefore$ length of the pipe $P_2$ will be $20\ cm$

41   In a resonance column experiment, a tuning fork of frequency $400\ Hz$ is used. The first resonance is observed when the air column has a length of $20.0\ cm$ and the second resonance is observed when the air column has a length of $62.0\ cm$. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form ?

##### Solution :

Length of the wire $= 1.0\ m$ $\\$ For fundamental frequency $\lambda/2 = l$ $\\$ $\Rightarrow \lambda = 2l = 2 \times 1 = 2\ m$ $\\$ Here given $n = 3.8\ km/s = 3800\ m/s$ $\\$ We know $\Rightarrow v = n\lambda \Rightarrow n = \dfrac{3800}{2} = 1.9\ kH$. $\\$ So standing frequency between $20\ Hz$ and $20\ kHz$ which will be heard are $\\$ $= n \times 1.9\ kHz$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ where $n = 0, 1, 2, 3, … 10$

42   The first overtone frequency of a closed organ pipe $P_1$ is equal to the fundamental frequency of an open organ pipe $P_2$ . If the length of the pipe $P_1$ is $30\ cm$, what will be the length of $P_2$ ?

##### Solution :

Let the length will be $l$. $\\$ Here given that $V = 340\ m/s$ and $n = 20\ Hz$ $\\$ Here $\lambda/2 = l \Rightarrow \lambda = 2l$ $\\$ We know $V = n\lambda \Rightarrow l = \dfrac{V}{n} = \dfrac{340}{2 \times 20} = \dfrac{34}{4} = 8.5\ cm$ $\ \ \$ (for maximum wavelength, the frequency is minimum)

43   A copper rod of length $1\cdot0\ m$ is clamped at its middle point. Find the frequencies between $20\ Hz \ - \ 20,000\ Hz$ at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is $3.8\ km/s$.

##### Solution :

a) Here given $l = 5\ cm = 5 \times 10^{–2} m,\ v = 340\ m/s$ $\\$ $\Rightarrow n = \dfrac{V}{2I} = \dfrac{340}{2 \times 5 \times 10^{-2}} = 3.4\ kHz$ $\\$ b) If the fundamental frequency $= 3.4\ KHz$ $\\$ $\Rightarrow$ then the highest harmonic in the audible range $(20\ Hz – 20\ KHz)$ $\\$ $= \dfrac{20000}{3400} = 5.8 = 5$ $\ \ \ \ \ \ \$ (integral multiple of $3.4 KHz$).

44   Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range $(20 - 20,000\ Hz)$. Speed of sound in $air = 340\ m/s$.

##### Solution :

The resonance column apparatus is equivalent to a closed organ pipe. $\\$ Here $I = 80\ cm = 10 \times 10^{–2}\ m ;\ v = 320\ m/s$ $\\$ $\Rightarrow n_{0} = \dfrac{v}{4I} = \dfrac{320}{4 \times 50 \times 10^{-2}} = 100\ Hz$ $\\$ So the frequency of the other harmonics are odd multiple of $n_{0} = (2n + 1)\ 100\ Hz$ $\\$ According to the question, the harmonic should be between $20\ Hz$ and $2\ KHz$.

45   An open organ pipe has a length of $5\ cm$. (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highest harmonic of such a tube that is in the audible range ? Speed of sound in air is $340\ m/s$ and the audible range is $20 - 20,000\ Hz$.

##### Solution :

Let the length of the resonating column will be $= 1$ $\\$ Here $V = 320\ m/s$ $\\$ Then the two successive resonance frequencies are $\dfrac{(n + 1)v}{4I}$ and $\dfrac{nv}{4I}$ $\\$ Here given $\dfrac{(n + 1)v}{4I} = 2592 ;\ \lambda = \dfrac{nv}{4I} = 1944$ $\\$ $\Rightarrow \dfrac{(n + 1)v}{4I} - \dfrac{nv}{4I} = 2592 - 1944 = 548\ cm = 25\ cm.$

46   An electronically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of air column in the tube is $80\ cm$. The frequency of the loudspeaker can be varied between $20\ Hz - 2\ kHz$. Find the frequencies at which the column will resonate. Speed of sound in air $= 320\ m/s$.

##### Solution :

Let, the piston resonates at length $l_{1}$ and $l_{2}$ $\\$ Here, $l = 32\ cm;\ v = ?, n = 512\ Hz$ $\\$ Now $\Rightarrow 512 = v/\lambda$ $\\$ $\Rightarrow v = 512 \times 0.64 = 328\ m/s.$

47   Two successive resonance frequencies in an open organ pipe are $1944\ Hz$ and $2592\ Hz$. Find the length of the tube. The speed of sound in air is $324\ m/s$.

##### Solution :

Let the length of the longer tube be $L_2$ and smaller will be $L_1$. $\\$ According to the data $440 = \dfrac{3 \times 330}{4 \times L_{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .............(1)$ (first over tone) $\\$ And $440 = \dfrac{330}{4 \times L_{1}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..............(2)$ (fundamental) $\\$ solving equation we get $L_{2} = 56.3\ cm$ and $L_{1} = 18.8\ cm$.

48   A piston is fitted in a cylindrical tube of small cross-section with the other end of the tube open. The tube resonates with a tuning fork of frequency $512\ Hz.$ The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of $32\cdot0\ cm$. Calculate the speed of sound in the air of the tube.

##### Solution :

Let $n_{0}$ = frequency of the turning fork, $T$ = tension of the string $\\$ $L = 40 cm = 0.4\ m,\ m = 4g = 4 \times 10^{–3}\ kg$ $\\$ So, $m$ = Mass/Unit length $= 10^{–2}\ kg/m$ $\\$ $n_{0} = \dfrac{1}{21}\sqrt{\dfrac{T}{m}}$ $\\$ So, $2^{nd}$ harmonic $2n_{0} = \Big(\dfrac{2}{2I}\Big)\sqrt{\dfrac{T}{m}}$ $\\$ As it is unison with fundamental frequency of vibration in the air column $\\$ $\Rightarrow 2n_{0} = \dfrac{340}{4 \times 1} = 85\ Hz$ $\\$ $\Rightarrow 85 = \dfrac{2}{2 \times 0.4}\sqrt{\dfrac{T}{14}} \Rightarrow T = 85^{2} \times (0.4)^2 \times 10^{-2} = 11.6$ Newton.

49   A U-tube having unequal arm-lengths has water in it. A tuning fork of frequency $440\ Hz$ can set up the air in the shorter arm in its fundamental mode of vibration and the same tuning fork can set up the air in the longer arm in its first overtone vibration. Find the length of the air columns. Neglect any end effect and assume that the speed of sound in air $= 330\ m/s$.

##### Solution :

Given, $m = 10\ g = 10 \times 10^{–3} kg,\ l = 30\ cm = 0.3\ m$ $\\$ Let the tension in the string will be $= T$ $\\$ $\mu$ = mass / unit length $= 33 \times 10^{–3}\ kg$ $\\$ The fundamental frequency $\Rightarrow n_{0} = \dfrac{1}{2I}\sqrt{\dfrac{T}{\mu}} \ \ \ \ \ \ \ \ \ \ \ \ \ .......(1)$ $\\$ The fundamental frequency of closed pipe $\\$ $\Rightarrow n_{0} = \Big(\dfrac{v}{4I}\Big) \dfrac{340}{4 \times 50 \times 10^2} = 170\ Hz \ \ \ \ \ \ \ \ \ \ \ \ \ .....(2)$ $\\$ According to equations (1) \times (2) we get, $\\$ $170 = \dfrac{1}{2 \times 30 \times 10^{-2}} \times \sqrt{\dfrac{T}{33 \times 10^{-3}}}$ $\\$ $\Rightarrow T = 347$ Newton

50   Consider the situation shown in figure (16-E9). The wire which has a mass of $4\cdot00\ g$ oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is $340\ m/s$, find the tension in the wire.

##### Solution :

We know that $f\ \infty\ \sqrt{T}$ $\\$ According to the question $f + \Delta{f}\ \infty\ \sqrt{\Delta{T}} + T$ $\\$ $\Rightarrow \dfrac{f + \Delta{f}}{f} = \sqrt{\dfrac{\Delta{t} + T}{T}} \Rightarrow 1 + \dfrac{\Delta{f}}{f} = \Big(1 + \dfrac{\Delta{T}}{T}\Big)^\dfrac{1}{2} = 1 + \dfrac{1}{2} \dfrac{\Delta{T}}{T} +........ \ \ \ \$ (neglecting other terms) $\\$ $\Rightarrow \dfrac{\Delta{f}}{f} = \Big(\dfrac{1}{2}\Big)\dfrac{\Delta{T}}{T}.$

51   A $30\cdot0-cm$-long wire having a mass of $10\cdot0\ g$ is fixed at the two ends and is vibrated in its fundamental mode. A $50\cdot0-cm$-long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air $= 340\ m/s$.

##### Solution :

We know that the frequency $= f,\ T$ = temperatures $\\$ $f\ \infty\ \sqrt{T}$ $\\$ So,$\dfrac{f_1}{f_2} = \dfrac{\sqrt{T_1}}{\sqrt{T_2}} \Rightarrow \dfrac{293}{f_2} = \dfrac{\sqrt{293}}{\sqrt{295}}$ $\\$ $\Rightarrow f_2 = \dfrac{293 \times \sqrt{295}}{\sqrt{293}} = 294$

52   Show that if the room temperature changes by a small amount from $T$ to $T\ +\ \Delta{T}$, the fundamental frequency of an organ pipe changes from $v$ to $v\ +\ \Delta{v}$, where $\\$ $$\dfrac{\Delta{v}}{v}=\dfrac{1}{2}\dfrac{\Delta{T}}{T}$$

##### Solution :

We know that $f\ \infty\ \sqrt{T}$ $\\$ According to the question $f + \Delta{f}\ \infty\ \sqrt{\Delta{T}} + T$ $\\$ $\Rightarrow \dfrac{f + \Delta{f}}{f} = \sqrt{\dfrac{\Delta{t} + T}{T}} \Rightarrow 1 + \dfrac{\Delta{f}}{f} = \Big(1 + \dfrac{\Delta{T}}{T}\Big)^\dfrac{1}{2} = 1 + \dfrac{1}{2} \dfrac{\Delta{T}}{T} +........ \ \ \ \$ (neglecting other terms) $\\$ $\Rightarrow \dfrac{\Delta{f}}{f} = \Big(\dfrac{1}{2}\Big)\dfrac{\Delta{T}}{T}.$

53   The fundamental frequency of a closed pipe is $293\ Hz$ when the air in it is at a temperature of $20°C$. What will be its fundamental frequency when the temperature changes to $22°C$ ?

##### Solution :

54   A Kundt's tube apparatus has a copper rod of length $1\cdot0\ m$ clamped at $25\ cm$ from one of the ends. The tube contains air in which the speed of sound is $340\ m/s$. The powder collects in heaps separated by a distance of $5\cdot0\ cm$. Find the speed of sound waves in copper.

##### Solution :

55   A Kundt's tube apparatus has a steel rod of length $1\cdot0\ m$ clamped at the centre. It is vibrated in its fundamental mode at a frequency of $2600\ Hz$. The lycopodium powder dispersed in the tube collects into heaps separated by $6.5\ cm$. Calculate the speed of sound in steel and in air.

##### Solution :

56   A source of sound with adjustable frequency produces $2$ beats per second with a tuning fork when its frequency is either $476\ Hz$ or $480\ Hz$. What is the frequency of the tuning fork ?

##### Solution :

57   A tuning fork produces $4$ beats per second with another tuning fork of frequency $256\ Hz$. The first one is now loaded with a little wax and the beat frequency is found to increase to $6$ per second. What was the original frequency of the tuning fork ?

##### Solution :

58   Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength of $32\ cm$ and the other of $32\cdot2\ cm$. The speed of sound in air is $350\ m/s$.

##### Solution :

59   A tuning fork of unknown frequency makes $5$ beats per second with another tuning fork which can cause a closed organ pipe of length $40\ cm$ to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is $320\ m/s$.

##### Solution :

60   A piano wire $A$ vibrates at a fundamental frequency of $600\ Hz$. A second identical wire $B$ produces $6$ beats per second with it when the tension in $A$ is slightly increased. Find the ratio of the tension in $A$ to the tension in $B$.

##### Solution :

61   A tuning fork of frequency $256\ Hz$ produces $4$ beats per second with a wire of length $25\ cm$ vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork ?

##### Solution :

62   A traffic policeman standing on a road sounds a whistle emitting the main frequency of $2\cdot00\ kHz$. What could be the appparent frequency heard by a scooter-driver approaching the policeman at a speed of $36\cdot0\ km/h$ ? Speed of sound in air $= 340\ m/s$.

##### Solution :

63   The horn of a car emits sound with a dominant frequency of $2400\ Hz$. What will be the apparent dominant frequency heard by a person standing on the road in front of the car if the car is approaching at $18\cdot0\ km/h$ ? Speed of sound in air $= 340\ m/s$.

##### Solution :

64   A person riding a car moving at $72\ km/h$ sounds a whistle emitting a wave of frequency $1250\ Hz$. What frequency will be heard by another person standing on the road (a) in front of the car (b) behind the car ? Speed of sound in air $= 340\ m/s.$

##### Solution :

65   A train approaching a platform at a speed of $54\ km/h$ sounds a whistle. An observer on the platform finds its frequency to be $1620\ Hz$. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platfrom ? The speed of sound in air $= 332\ m/s$.

##### Solution :

66   A bat emitting an ultrasonic wave of frequency $4.5 \times 10^4\ Hz$ flies at a speed of $6\ m/s$ between two parallel walls. Find the two frequencies heard by the bat and the beat frequency between the two. The speed of sound is $330\ m/s$.

##### Solution :

67   A bullet passes past a person at a speed of $220\ m/s$. Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. Speed of sound in air $= 330\ m/s$.

##### Solution :

68   Two electric trains run at the same speed of $72\ km/h$ along the same track and in the same direction with a separation of $2.4\ km$ between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of $500\ m$ from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at $500\ Hz$ and the speed of sound in air is $340\ m/s$, find the frequencies heard by the person

##### Solution :

69   A violin player riding on a slow train plays a $440\ Hz$ note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears $4\cdot0$ beats per second. The speed of sound in air $= 340\ m/s$. (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train ?

##### Solution :

70   Two identical tuning forks vibrating at the same frequency $256\ Hz$ are kept fixed at some distance apart. A listener runs between the forks at a speed of $3\cdot0\ m/s$ so that he approaches one tuning fork and recedes from the other (figure 16-E10). Find the beat frequency observed by the listener. Speed of sound in air $= 332\ m/s$.

##### Solution :

71   Figure (16-E11) shows a person standing somewhere in between two identical tuning forks, each vibrating at $512\ Hz.$ If both the tuning fbrks move towards right at a speed of $5\cdot5\ m/s$, find the number of beats heard by the listener. Speed of sound in air $= 330\ m/s$.

##### Solution :

72   A small source of sound vibrating at frequency $500\ Hz$ is rotated in a circle of radius $100/\pi\ cm$ at a constant angular speed of $5\cdot0$ revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and the maximum frequency of the sound observed. Speed of sound in air $= 332\ m/s$.

##### Solution :

73   Two trains are travelling towards each other both at a speed of $90\ km/h$. If one of the trains sounds a whistle at $500\ Hz,$ what will be the apparent frequency heard in the other train ? Speed of sound in air $= 350\ m/s$.

##### Solution :

74   A traffic policeman sounds a whistle to stop a car-driver approaching towards him. The car-driver does not stop and takes the plea in court that because of the Doppler shift, the frequency of the whistle reaching him might have gone beyond the audible limit of $20\ kHz$ and he did not hear it. Experiments showed that the whistle emits a sound with frequency close to $16\ kHz$. Assuming that the claim of the driver is true, how fast was he driving the car ? Take the speed of sound in air to be $330\ m/s$. Is this speed practical with today's technology ?

##### Solution :

75   A car moving at $108\ km/h$ finds another car in front of it going in the same direction at $72\ km/h$. The first car sounds a horn that has a dominant frequency of $800\ Hz$. What will be the apparent frequency heard by the driver in the front car ? Speed of sound in air $= 330\ m/s$.

##### Solution :

76   Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of $36\ km/h$ and the other at $54\ km/h$ relative to the water. The first submarine sends a sound signal (sound waves in water are also called sonar) at a frequency of $2000\ Hz.$ (a) At what frequency is this signal received by the second submarine ? (b) The signal is reflected from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of the sound wave in water to be $1500\ m/s$.

##### Solution :

77   A small source of sound oscillates in simple harmonic motion with an amplitude of $17\ cm$. A detector is placed along the line of motion of the source. The source emits a sound of frequency $800\ Hz$ which travels at a speed of $340\ m/s$. If the width of the frequency band detected by the detector is $8\ Hz$, find the time period of the source.

##### Solution :

78   A boy riding on his bike is going towards east at a speed. of $4\ \sqrt{2}\ m/s$. At a certain point he produces a sound pulse of frequency $1650\ Hz$ that travels in air at a speed of $334\ m/s$. A second boy stands on the ground $45°$ south of east from him. Find the frequency of the pulse as received by the second boy.

##### Solution :

79   A sound source, fixed at the origin, is continuously emitting sound at a frequency of $660\ Hz$. The sound travels in air at a speed of $330\ m/s$. A listener is moving along the line $x = 336\ m$ at a constant speed of $26\ m/s$. Find the frequency of the sound as observed by the listener when he is (a) at $y = — 140\ m$, (b) at $y = 0$ and (c) at $y = 140\ m$.