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:
Where
v = speed of sound
$\gamma$= adiabatic constant (ratio of specific heats) for air (approximately 1.4)
P = pressure of the air
ρ = density of the air
(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)
$\frac{v_{m}}{v_{d}} = \sqrt{\frac{\frac{\gamma P}{\rho_{m}}} {\frac{\gamma P}{\rho_{d}}}}$
$\frac{v_{m}}{v_{d}} = \sqrt{\frac{\gamma P}{\rho_{m}} \times \frac{\rho_{d}}{\gamma P}}$
$\frac{v_{m}}{v_{d}} =\sqrt{ \frac{ \rho_{d}}{\rho_{m}}}$
Effect of Humidity: When humidity increases, the air contains more water vapor molecules.
* Molar Mass: Water vapor has a lower molar mass than dry air.
* Density Reduction: For a given pressure and temperature, air with higher humidity will have a slightly lower density.
$\rho_{d} < \rho_{m}$
$\therefore v_{m} > v_{d}$
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 learned that a traveling wave in one dimension is represented by a function y = f(x, t) where x and t must appear in the combination x - vt or x + vt, i.e., y = f(x ± vt). Is the converse true? Examine if the following functions for y can possibly represent a traveling wave:
(a) y = (x - vt)²
(b) y = log[(x + vt)/x₀]
(c) y = 1/(x + vt)
Solution:
Understanding Traveling Waves
A traveling wave in one dimension is characterized by a function that maintains its shape as it moves along the x-axis with a constant speed (v). This is typically represented by:
y = f(x ± vt)
where:
* y: Wave displacement
* x: Position
* t: Time
* v: Wave speed
* ±: Direction of wave propagation (positive for rightward, negative for leftward)
Analyzing the Given Functions
Let's examine each function to see if it fits the form of a traveling wave:
(a) y = (x - vt)²
* This function does not represent a traveling wave. The variable 't' is not combined with 'x' in the form 'x ± vt'.
(b) y = log[(x + vt)/x₀]
* This function does represent a traveling wave. The argument of the logarithm, (x + vt)/x₀, is a function of (x + vt). This indicates that the wave is propagating to the left with speed 'v'.
(c) y = 1/(x + vt)
* This function does represent a traveling wave. The denominator, (x + vt), is a function of (x + vt). This suggests that the wave is propagating to the left with speed 'v'.
Conclusion
Therefore, only functions (b) and (c) can possibly represent traveling waves, as they satisfy the condition of having 'x' and 't' combined in the form '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((2π/3)x) cos(120πt)
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0 x 10^-2$ kg.
Answer the following:
(a) Does the function represent a traveling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves traveling in opposite directions. What is the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.
14.12 (i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?
14.13 Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all:
(a) y = 2 cos (3x) sin (10t)
(b) y= 2√x - ut
(c)y = 3 sin (5x0.5t) + 4 cos (5x-0.5t)
(d)y = cos x sin t + cos 2x sin 2t
14.14 A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10-2 kg and its linear mass density is 4.0 x 10-2 kg m-¹. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
14.15 A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
14.16 A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
14.17 A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s¯¹)
14.18 Two sitar strings A and B playing the note 'Ga' are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
14.19 Explain why (or how):
(a)in a sound wave, a displacement node is a pressure antinode and vice versa,
(b) bats canca violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d)solids can support both longitudinal and transverse waves, but only
longitudinal waves can propagate in gases, and
(e) the shape of a pulse gets distorted during propagation in a dispersive medium.