Q: What is the speed of sound in (i) a steel rod (ii) water? Young's modulus of steel is $20 \times 10^{10}\text{ N m}^{-2}$ and density of steel is $7.8 \times 10^3\text{ kg m}^{-3}$. The bulk modulus of water is $0.22 \times 10^{10}\text{ N m}^{-2}$ and density of water is $10^3\text{ kg m}^{-3}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans:
(i) $Y = 20 \times 10^{10}\text{ N m}^{-2}$, $\rho = 7.8 \times 10^3\text{ kg m}^{-3}$
$v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20 \times 10^{10}}{7.8 \times 10^3}} = \mathbf{5.06 \times 10^3\text{ m s}^{-1}\text{ (or) } 5063\text{ m s}^{-1}}$
(ii) $B = 0.22 \times 10^{10}\text{ N m}^{-2}$, $\rho = 10^3\text{ kg m}^{-3}$
$v = \sqrt{\frac{B}{\rho}} = \sqrt{\frac{0.22 \times 10^{10}}{10^3}} = \mathbf{1.48 \times 10^3\text{ m s}^{-1}\text{ (or) } 1483\text{ m s}^{-1}}$
Q: The speed of sound in a particular liquid is $1.6 \times 10^3\text{ m s}^{-1}$. If the bulk modulus of the liquid is $0.15 \times 10^{10}\text{ N m}^{-2}$, what is the density of the liquid?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $B = 0.15 \times 10^{10}\text{ N m}^{-2}$, $v = 1.6 \times 10^3\text{ m s}^{-1}$, $\rho = ?$
$\rho = \frac{B}{v^2} = \frac{0.15 \times 10^{10}}{(1.6 \times 10^3)^2} = \mathbf{585.94\text{ kg m}^{-3}}$
Q: Calculate the velocity of sound in air at STP. The density of air at STP is $1.29\text{ kg m}^{-3}$. Assume air to be diatomic with $\gamma = 1.4$. Also calculate the velocity of sound in air at $27^\circ\text{C}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: At STP: $p = 1.01 \times 10^5\text{ Pa}$, $T_1 = 273\text{ K}$, $\rho = 1.29\text{ kg m}^{-3}$, $\gamma = 1.4$
$v_1 = \sqrt{\frac{\gamma p}{\rho}} = \sqrt{\frac{1.4 \times 1.01 \times 10^5}{1.29}} = \mathbf{331.077\text{ m s}^{-1}\text{ (at STP)}}$
At $27^\circ\text{C}$ ($T_2 = 27 + 273 = 300\text{ K}$):
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \implies v_2 = v_1 \sqrt{\frac{300}{273}} = 331.077 \times \sqrt{\frac{300}{273}} = \mathbf{347.06\text{ m s}^{-1}}$
Q: At a busy street corner, the intensity level of sound is $70\text{ dB}$. What is intensity of sound in $\text{W m}^{-2}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $\beta = 70\text{ dB}$, $I_0 = 10^{-12}\text{ W m}^{-2}$
$\beta = 10 \log_{10}\left(\frac{I}{I_0}\right) \implies 70 = 10 \log_{10}\left(\frac{I}{10^{-12}}\right)$
$\frac{I}{10^{-12}} = 10^7 \implies I = 10^{-12} \times 10^7 = \mathbf{10^{-5}\text{ W m}^{-2}}$
Q: What value of sound intensity in $\text{dB}$ increases by a factor of $1000$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $\frac{I_2}{I_1} = 1000$, $\beta = ?$
$\beta = 10 \log_{10}\left(\frac{I_2}{I_1}\right) = 10 \log_{10}(1000) = 10 \times 3 = \mathbf{30\text{ dB}}$
The sound level will increase $30\text{ dB}$.
Q: If the intensity of sound is doubled, by how many decibels does the sound level increase?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $\frac{I_2}{I_1} = 2$, $\beta = ?$
$\beta = 10 \log_{10}(2) = 10 \times 0.3010 = \mathbf{3\text{ dB}}$
The sound level will increase $3\text{ dB}$.
Q: A sonar echo returns to a submarine $1.2\text{ s}$ after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water and the velocity of sound in water is $1500\text{ m s}^{-1}$.)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $t = 1.2\text{ s}$, $v = 1500\text{ m s}^{-1}$, $d = ?$
$d = \frac{v t}{2} = \frac{1500 \times 1.2}{2} = \mathbf{900\text{ m}}$
Q: A bat flying at a speed of $20\text{ m s}^{-1}$ towards a stationary moth emits sound wave of $340\text{ m s}^{-1}$. The bat hears the echoes $0.4\text{ s}$ later. (i) Calculate the distance travelled by the bat in $0.4\text{ s}$. (ii) Calculate the distance travelled by the sound wave in $0.4\text{ s}$. (iii) Determine the initial distance between the bat and the moth.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v_b = 20\text{ m s}^{-1}$, $v_s = 340\text{ m s}^{-1}$, $t = 0.4\text{ s}$
(i) Distance travelled by bat ($d_b$) $= v_b t = 20 \times 0.4 = \mathbf{8\text{ m}}$
(ii) Distance travelled by sound wave ($d_s$) $= v_s t = 340 \times 0.4 = \mathbf{136\text{ m}}$
(iii) Initial distance $= \frac{d_s + d_b}{2}$ or total distance / 2 $= \frac{8 + 136}{2} = \mathbf{72\text{ m}}$
Q: The sole survivor of a shipwreck swim to an island, which is $3000\text{ m}$ from a vertical cliff. He sees a ship anchored between the island and the cliff. A blast from the ship horn is heard twice with the time elapse of $4\text{ s}$. Calculate distance between the ship and the survivor. (Speed of sound in air = $330\text{ m s}^{-1}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: Distance between survivor and ship $= d$
Distance between survivor and cliff $= 3000\text{ m}$, Distance between ship and cliff $= 3000 - d$
$v = 330\text{ m s}^{-1}$, $t_2 - t_1 = 4\text{ s}$
Time for original sound ($t_1$) $= \frac{d}{v}$
Time for echo ($t_2$) $= \frac{3000 + (3000 - d)}{v} = \frac{6000 - d}{v}$
$t_2 - t_1 = \frac{6000 - d}{v} - \frac{d}{v} \implies 4 = \frac{6000 - 2d}{330}$
$1320 = 6000 - 2d \implies 2d = 4680 \implies d = \mathbf{2340\text{ m}}$
Q: A piezoelectric transducer produces ultrasound at a frequency of $10\text{ MHz}$. Calculate the wavelength of this ultrasound in blood if the speed of ultrasound in blood is $1560\text{ m s}^{-1}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $f = 10\text{ MHz} = 10 \times 10^6\text{ Hz}$, $v = 1560\text{ m s}^{-1}$, $\lambda = ?$
$\lambda = \frac{v}{f} = \frac{1560}{10 \times 10^6} = \mathbf{1.56 \times 10^{-4}\text{ m}}$
Q: (i) Why the ultrasound is used in medical imaging rather than sound wave of audible frequencies or lower? (ii) The frequencies used in imaging are typically in the range of $1\text{ to } 15\text{ MHz}$. What is the range of wavelengths in human tissue where the velocity of sound is $1540\text{ m s}^{-1}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans:
(i) Because ultrasound has higher frequency and shorter wavelength. Shorter wavelengths are more easily reflected or refracted in the superficial tissues than longer wavelengths, giving better resolution.
(ii) $v = 1540\text{ m s}^{-1}$, $f_1 = 1\text{ MHz} = 10^6\text{ Hz}$, $f_2 = 15\text{ MHz} = 15 \times 10^6\text{ Hz}$
$\lambda_1 = \frac{v}{f_1} = \frac{1540}{1 \times 10^6} = 1.54 \times 10^{-3}\text{ m}$
$\lambda_2 = \frac{v}{f_2} = \frac{1540}{15 \times 10^6} = 0.103 \times 10^{-3}\text{ m}$
The range of wavelengths is between $\mathbf{0.103 \times 10^{-3}\text{ m} \text{ and } 1.54 \times 10^{-3}\text{ m}}$.
Q: The velocity of ultrasound in lung (inclusive air), fat and skull bone are $600\text{ m s}^{-1}$, $1450\text{ m s}^{-1}$ and $4080\text{ m s}^{-1}$ respectively, what are the wavelengths of ultrasound if the sound waves at $2\text{ MHz}$ are used?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $f = 2\text{ MHz} = 2 \times 10^6\text{ Hz}$
- In lung ($v_1 = 600\text{ m s}^{-1}$): $\lambda_1 = \frac{v_1}{f} = \frac{600}{2 \times 10^6} = \mathbf{3 \times 10^{-4}\text{ m}}$
- In fat ($v_2 = 1450\text{ m s}^{-1}$): $\lambda_2 = \frac{v_2}{f} = \frac{1450}{2 \times 10^6} = \mathbf{7.25 \times 10^{-4}\text{ m}}$
- In skull ($v_3 = 4080\text{ m s}^{-1}$): $\lambda_3 = \frac{v_3}{f} = \frac{4080}{2 \times 10^6} = \mathbf{2.04 \times 10^{-3}\text{ m}}$
Q: To get the better resolution, higher frequencies ultrasound are used but at the expense of less penetration because sound waves are attenuated within the distance about $50\lambda$ in tissue. If the velocity of sound in soft tissue is $1500\text{ m s}^{-1}$ and mean depth of thyroid is $0.75\text{ cm}$. What is the most suitable frequency of ultrasound to diagnose thyroid?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v = 1500\text{ m s}^{-1}$, $50\lambda = \text{mean depth} = 0.75\text{ cm} = 0.75 \times 10^{-2}\text{ m}$
$\lambda = \frac{0.75 \times 10^{-2}}{50} = 1.5 \times 10^{-4}\text{ m}$
$f = \frac{v}{\lambda} = \frac{1500}{1.5 \times 10^{-4}} = \mathbf{10 \times 10^6\text{ Hz} = 10\text{ MHz}}$
Q: An ambulance travelling at $44\text{ m s}^{-1}$ approaches a car heading in a same direction at the speed of $28\text{ m s}^{-1}$. The ambulance driver has a siren sounding at $550\text{ Hz}$. At what frequency does the driver of the car hear the siren? (Speed of siren sound $= v = 330\text{ m s}^{-1}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: For ambulance (source): $f_s = 550\text{ Hz}$, $v_s = 44\text{ m s}^{-1}$
For car (observer): $v_o = 28\text{ m s}^{-1}$, $v = 330\text{ m s}^{-1}$
$f_o = f_s \left(\frac{v - v_o}{v - v_s}\right) = 550 \left(\frac{330 - 28}{330 - 44}\right) = 550 \left(\frac{302}{286}\right) = \mathbf{580.77\text{ Hz}}$
Q: In example 7, what will be the frequency of the siren heard by the driver of the car when the ambulance overtakes his car?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: For ambulance (source): $f_s = 550\text{ Hz}$, $v_s = 44\text{ m s}^{-1}$
For car (observer): $v_o = 28\text{ m s}^{-1}$, $v = 330\text{ m s}^{-1}$
$f_o = f_s \left(\frac{v + v_o}{v + v_s}\right) = 550 \left(\frac{330 + 28}{330 + 44}\right) = 550 \left(\frac{358}{374}\right) = \mathbf{526.47\text{ Hz}}$
Q: A monorail approaches a platform at a speed of $50\text{ m s}^{-1}$ which it blows its whistle. A man standing on the platform hears the whistle with frequency $261\text{ Hz}$. There is no wind and temperature is chilly $0^\circ\text{C}$. What is the observed frequency of the whistle when the train is at rest? (The speed of sound in air at $0^\circ\text{C}$ is $332\text{ m s}^{-1}$.)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: For man (observer): $f_o = 261\text{ Hz}$, $v_o = 0$, $v = 332\text{ m s}^{-1}$
For train (source): $v_s = 50\text{ m s}^{-1}$, $f_s = ?$
$f_o = f_s \left(\frac{v}{v - v_s}\right) \implies 261 = f_s \left(\frac{332}{332 - 50}\right) = f_s \left(\frac{332}{282}\right)$
$f_s = \frac{261 \times 282}{332} = \mathbf{221.69\text{ Hz}}$
Q: A car is travelling at $29\text{ m s}^{-1}$ towards a stationary whistle with a frequency of $625\text{ Hz}$. If the speed of sound is $337\text{ m s}^{-1}$, what is the apparent frequency of the whistle as heard by the driver of the car?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: For whistle (source): $v_s = 0$, $f_s = 625\text{ Hz}$
For car driver (observer): $v_o = 29\text{ m s}^{-1}$, $v = 337\text{ m s}^{-1}$
$f_o = f_s \left(\frac{v + v_o}{v}\right) = 625 \left(\frac{337 + 29}{337}\right) = 625 \left(\frac{366}{337}\right) = \mathbf{679\text{ Hz}}$
Q: (i) What frequency is received by a person watching an oncoming ambulance moving at $110\text{ km h}^{-1}$ and emitting a steady $800\text{ Hz}$ sound from its siren? The speed of sound on this day is $345\text{ m s}^{-1}$. (ii) What frequency does she receive after the ambulance has passed?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: For ambulance (source): $f_s = 800\text{ Hz}$, $v_s = 110\text{ km h}^{-1} = 110 \times \frac{10^3}{3600} = 30.56\text{ m s}^{-1}$
For person (observer): $v_o = 0$, $v = 345\text{ m s}^{-1}$
(i) Approaching ($-v_s$):
$f_o = f_s \left(\frac{v + v_o}{v - v_s}\right) = 800 \left(\frac{345 + 0}{345 - 30.56}\right) = \mathbf{877.75\text{ Hz}}$
(ii) Receding ($+v_s$):
$f_o = f_s \left(\frac{v + v_o}{v + v_s}\right) = 800 \left(\frac{345 + 0}{345 + 30.56}\right) = \mathbf{734.9\text{ Hz}}$
Q: A $5000\text{ Hz}$ sound wave is emitted by a stationary source. This sound reflects from an object moving $3.5\text{ m s}^{-1}$ toward the source. What is the frequency of the wave reflected by moving object as detected by a detector at rest near the source?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: Step 1: Moving object acts as observer moving towards stationary source.
$f_s = 5000\text{ Hz}$, $v_s = 0$, $v_o = 3.5\text{ m s}^{-1}$, $v = 345\text{ m s}^{-1}$
$f_o = f_s \left(\frac{v + v_o}{v - v_s}\right) = 5000 \left(\frac{345 + 3.5}{345 + 0}\right) = 5050.7\text{ Hz}$
Step 2: Moving object acts as source ($f_s' = 5050.7\text{ Hz}$, $v_s' = 3.5\text{ m s}^{-1}$) and detector is at rest ($v_o' = 0$).
$f_o' = f_s' \left(\frac{v + v_o'}{v - v_s'}\right) = 5050.7 \left(\frac{345 + 0}{345 - 3.5}\right) = \mathbf{5102.5\text{ Hz}}$
Q: Calculate the displacement of air molecules (amplitude) for a sound having a frequency of $1000\text{ Hz}$ at the threshold of hearing. The speed of sound in air is $343\text{ m s}^{-1}$ at $20^\circ\text{C}$ and density of air is $1.29\text{ kg m}^{-3}$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $f = 1000\text{ Hz}$, $v = 343\text{ m s}^{-1}$, $\rho = 1.29\text{ kg m}^{-3}$, $I = 1 \times 10^{-12}\text{ W m}^{-2}$
$\omega = 2\pi f = 2 \times 3.142 \times 1000 = 6284\text{ rad s}^{-1}$
$I = \frac{1}{2} \rho v \omega^2 A^2 \implies 1 \times 10^{-12} = \frac{1}{2} \times 1.29 \times 343 \times (6284)^2 \times A^2$
$A^2 = \frac{2 \times 10^{-12}}{1.29 \times 343 \times (6284)^2}$
$A = \mathbf{10.7 \times 10^{-12}\text{ m}}$