14.1 A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Solution:
$M \ of \ the \ string , M = 2.50 kg$
$Tension \ in \ the \ string , T = 200N$
$Length \ of \ the \ string , l = 20.0 m$
$Mass \ per \ unit \ length$
$\mu = \frac{M}{l}$
$\mu = \frac{2.50}{20}$
$\mu = 0.125 kg/m$
The velocity (v) of the transverse wave in the string is given by the relation:
$v= \sqrt{\frac{T}{\mu}}$
$v = \sqrt{\frac{200}{0.125}}$
$v = \sqrt{\frac{200 \times 1000}{125}}$
$v = \sqrt{1600}$
$v = 40 m/s$
Time taken by the disturbance to reach the other end ,
$t =\frac{l}{v}$
$t = \frac{20}{40}$
$t = 0.50 sec$
14.2 A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1 ? (g = 9.8 m s–2)
Solution:
Here's how to calculate the time taken for the splash to be heard at the top of the tower:
* Calculate the time taken for the stone to fall:
Use the second equation of motion:
$s = ut + \frac{1}{2}gt^{2}$
where:
* s = 300 m (height of the tower)
* u = 0 m/s (initial velocity of the stone)
* g = 9.8 m/s² (acceleration due to gravity)
* $t_{1}$= time taken for the stone to fall
Substituting the values:
$300 = 0 \times t_{1}+ \frac{1}{2} \times 9.8 \times t^{2}$
$300 = \frac{1}{2} \times 9.8 \times t_{1}^{2}$
$600 = 9.8 \times t_{1}^{2}$
$t_{1}^{2}= \frac{600}{9.8}$
$t_{1}= \sqrt{\frac{600}{9.8}}$
$t_{1}=\sqrt{61.22}$ seconds
$t_{1}=7.82$ seconds
* Calculate the time taken for the sound to travel back up:
Use the formula: time = distance / speed
where:
* distance = 300 m (height of the tower)
* speed = 340 m/s (speed of sound)
Substituting the values:
time $t_{2} = \frac{300}{340}$
time $t_{2} = 0.88$ seconds
* Calculate the total time:
The total time is the sum of the time taken for the stone to fall and the time taken for the sound to travel back up:
Total time $t= t_{1}+t_{2}$
Total time $t= 7.82 + 0.88$
total time $t=8.70$ seconds
Therefore, the splash is heard at the top of the tower after approximately 8.70 seconds.
14.3 A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1.
Solution:
The speed of a transverse wave on a string is given by:
$v = \sqrt{\frac{T}{\mu}}$
where:
* v is the speed of the wave
* T is the tension in the string
* μ is the linear mass density of the string
We can find the linear mass density (μ) by dividing the total mass (m) by the length (L):
$\mu= \frac{m}{L}$
Substituting the given values:
μ = 2.10 kg / 12.0 m
μ = 0.175 kg/m
Now, we need to find the tension (T) that will make the speed of the wave equal to the speed of sound in air (343 m/s).
Substituting the given values into the equation for the speed of the wave:
$v = \sqrt{\frac{T}{\mu}}$
$343 = \sqrt{\frac{T}{0.175}}$
Squaring both sides:
$(343)^{2}= \frac{T}{0.175}$
$T = (343)^{2} \times 0.175$
$T= 117649 \times 0.175$
$T = 20588.575N$
$T = 2.058 \times 10^{4}N$
$T = 2.06 \times 10^{4}N$
Therefore, the tension in the wire should be approximately 20620 Newtons to make the speed of a transverse wave on the wire equal to the speed of sound in dry air at 20 °C.
14.4 Use the formula $v = \sqrt{\frac{\gamma P}{\rho}}$
to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Solution:
(a) is independent of pressure,
$v = \sqrt{\frac{\gamma P}{\rho}}$ .......(i)
Where ,
Density , $\rho= \frac{Mass}{Volume} = \frac{M}{V}$
M = Molecular weight of the gas
V = Volume of the gas
Hence , equation (i) reduces to :
$v = \sqrt{\frac{\gamma PV}{M}}$ ......(ii) Now from the ideal gas equation for n= 1
PV = RT
For constant T , PV = Constant
Since both M and $\gamma$ are constant.
$v = \sqrt{\frac{\gamma RT}{M}}$
Hence , at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.
(b) increases with temperature
$v = \sqrt{\frac{\gamma RT}{M}}$ .........(iii)
Where ,
Mass , $M = \rho V$ is a constant
$\gamma$ and R are also constants
We conclude from the equation (iii) that
$v \propto \sqrt{T}$
Hence , the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e the speed of the sound increase with an increase in the temperature of the gaseous medium and vice versa.
(c) increases with humidity.
Let $v_{m}$ and $v_{d}$ be the speeds of sound in moist air and dry air respectively
Let $\rho_{m}$ and $\rho_{d}$ be the densities of moist air and dry air respectively.
Take the relation
$v = \sqrt{\frac{\gamma P}{\rho}}$
Hence , the speed of sound in moist air is
$v_{m}= \sqrt{\frac{\gamma P}{\rho_{m}}}$ ........(i)
And the speed of sound in dry air is
$v_{d}= \sqrt{\frac{\gamma P}{\rho_{d}}}$ ........(ii)
On dividing equation (i) and (ii)
$\frac{v_{m}}{v_{d}} = \sqrt{\frac{\frac{\gamma P}{\rho_{m}}} {\frac{\gamma P}{\rho_{d}}}}$
$\frac{v_{m}}{v_{d}} = \sqrt{\frac{\gamma P}{\rho_{m}} \times \frac{\rho_{d}}{\gamma P}}$
$\frac{v_{m}}{v_{d}} =\sqrt{ \frac{ \rho_{d}}{\rho_{m}}}$
However, the presence of water vapour reduces the density of air
$\rho_{d} < \rho_{m}$
$\therefore v_{m} > v_{d}$
Hence , the speed of sound in moist air is greater than it is in dry air. Thus , in a gaseous medium the speed of sound increase with humidity.
14.5 You have learnt that a travelling wave in one dimension is represented by a function
y = f (x, t) where x and t must appear in the combination x – v t or x + v t, i.e.
y = f (x ± v t). Is the converse true? Examine if the following functions for y can
possibly represent a travelling wave :
(a) (x – vt )
2
(b) log [(x + vt)/x0
]
(c) 1/(x + vt)
14.6 A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m/s and in water 1486 m/s.
Solution:
Understanding the Problem
We have a bat emitting ultrasonic sound waves in air. When these waves encounter a water surface, they are partially reflected and partially transmitted. We need to find the wavelengths of both the reflected and transmitted sound waves.
Diagram
Solution
(a) Reflected Sound
* Frequency: The frequency of the reflected sound remains the same as the incident sound, which is 1000 kHz.
* Wavelength: The wavelength of the reflected sound is the same as the wavelength of the incident sound in air.
* Calculation:
Wavelength
$\lambda = \frac{Speed \ (v)}{ Frequency \ (f)}$
$\lambda= \frac{340}{(1000 \times 10^{3} }$
$\lambda= 3.4 \times 10^{-4} m
$\lambda= 0.34 \times 10^{1} \times 10^{-4}m$
$\lambda = 0.34 \times 10^{-3}m$
$\lambda= 0.34 mm$
(b) Transmitted Sound
* Frequency: The frequency of the transmitted sound remains the same as the incident sound, which is 1000 kHz.
* Wavelength: The wavelength of the transmitted sound changes as it enters the water due to the change in the speed of sound.
* Calculation:
$\lambda= \frac{v}{f}$
$\lambda= \frac{1486}{1000 \times 10^{3}}$
$\lambda = 1.486 \times 10^{-3} m $
$\lambda= 1.486 mm$
Therefore, the wavelength of the reflected sound is 0.34 mm, and the wavelength of the transmitted sound is 1.486 mm.
14.7 A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1 ? The operating frequency of the scanner is 4.2 MHz.
Solution:
The speed of sound in the tissue is 1.7 km s-1 and the operating frequency of the scanner is 4.2 MHz. We need to find the wavelength of sound in the tissue.
The formula for wavelength is:
wavelength = speed of sound / frequency
Substituting the given values,
we get:
$\lambda= \frac{1.7 \ km/s}{4.2 \ MHz}$
$ \lambda= \frac{1.7 \times 10^{3}m/s}{4.2 \times 10^{6}Hz}$
$ \lambda= 4.04 /times 10^{-4}m$
Therefore, the wavelength of sound in the tissue is $4.04 /times 10^{-4}m$
14.8 A transverse harmonic wave on a string is described by
y(x, t) = 3.0 sin (36 t + 0.018 x + π/4)
where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave ?
If it is travelling, what are the speed and direction of its propagation ?
(b) What are its amplitude and frequency ?
(c) What is the initial phase at the origin ?
(d) What is the least distance between two successive crests in the wave ?
Solution:
y(x, t) = 3.0 sin (36 t + 0.018 x + π/4)
Compare this equation with
$y = A sin ( \omega t+kx + \phi)$
14.9 For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude,frequency or phase ?
14.10 For the travelling harmonic wave
y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35)
where x and y are in cm and t in s. Calculate the phase difference between oscillatory
motion of two points separated by a distance of
(a) 4 m,
(b) 0.5 m,
(c) λ/2,
(d) 3λ/4
14.11 The transverse displacement of a string (clamped at its both ends) is given by
y(x, t) = 0.06 sin
2
3
π
x
cos (120 πt)
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is
3.0 ×10–2 kg.
Answer the following :
(a) Does the function represent a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite
directions. What is the wavelength, frequency, and speed of each wave ?