Waves - Class 11 Physics - Chapter 14 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Waves | NCERT | Physics | Class 11
Consider two waves passing through the same string. The principle of superposition for displacement says that the net displacement of a particle on the string is the sum of the displacements produced by the two waves individually. Suppose we state similar principles for the net velocity of the particle and the net kinetic energy of the particle. Such a principle will be valid for:
A) Both the velocity and the kinetic energy.
B) The velocity but not for the kinetic energy.
C) The kinetic energy but not for the velocity.
D) Neither the velocity nor the kinetic energy.
The correct answer is Option B: The velocity but not for the kinetic energy.
According to the principle of superposition, the total displacement, $\vec{y}{\text{net}}$, of a particle on a string due to two waves is given by the vector sum of the displacements caused by each wave: $$ \vec{y}{\text{net}} = \vec{y}_1 + \vec{y}_2 $$
Here, $\vec{y}_1$ and $\vec{y}2$ are the displacement vectors produced by the two individual waves. To find the net velocity, we differentiate the displacement equation with respect to time: $$ \frac{d\vec{y}{\text{net}}}{dt} = \frac{d\vec{y}_1}{dt} + \frac{d\vec{y}2}{dt} $$ $$ \vec{v}{\text{net}} = \vec{v}_1 + \vec{v}_2 $$
Hence, the velocities also add vectorially, and the principle of superposition is applicable to net velocity.
To determine if this principle applies to kinetic energy, consider the kinetic energy, which is proportional to the square of the velocity. Squaring the net velocity: $$ v_{\text{net}}^2 = (\vec{v}_1 + \vec{v}2)^2 $$ Expanding this, we get: $$ v{\text{net}}^2 = v_1^2 + v_2^2 + 2\vec{v}_1 \cdot \vec{v}_2 $$
Since $v_{\text{net}}^2$ is not simply $v_1^2 + v_2^2$, but includes an additional cross-term, $2\vec{v}_1 \cdot \vec{v}_2$, it is evident that the principle of superposition does not apply to kinetic energy. This cross-term involves a product of velocities, indicating that kinetic energies do not simply add when two waves interact. Thus, any effect like interference (constructive or destructive) could alter the combined kinetic energy differently from merely summing them.
Therefore, the principle of superposition holds for the net velocity of the particle but not for the net kinetic energy in the context of waves on a string.
The equation of two light waves are $y_{1} = 6 \cos \omega t$, $y_{2} = 8 \cos (\omega t + \phi)$. The ratio of maximum to minimum intensities produced by the superposition of these waves will be
A) $49: 1$
B) $1 : 49$
C) $1 : 7$
D) $7 : 1$
Step 1: Formulate the Wave Equations
The equations of the light waves are given as: $$ y_1 = 6 \cos(\omega t) $$ $$ y_2 = 8 \cos(\omega t + \phi) $$
Step 2: Apply the Principle of Superposition
Key concept: When two waves superpose, the resultant wave's maximum amplitude is the sum of the individual waves' amplitudes, and the minimum amplitude is the absolute difference of these amplitudes.
If $y_1 = A_1 \cos(\omega t)$ and $y_2 = A_2 \cos(\omega t + \phi)$ represent the waves, then:
Maximum amplitude of resultant wave $(A_{max}) = A_1 + A_2$
Minimum amplitude of resultant wave $(A_{min}) = |A_1 - A_2|$
Step 3: Compute the Intensity Ratio
The intensity of a wave is proportional to the square of its amplitude, defined as: $$ I \propto A^2 $$
Calculation of Amplitudes:
Maximum Amplitude $(A_{max})$ = $6 + 8 = 14$
Minimum Amplitude $(A_{min})$ = $|6 - 8| = 2$
Calculation of Intensities:
Maximum Intensity $(I_{max}) = k \times (14^2) = 196k$
Minimum Intensity $(I_{min}) = k \times (2^2) = 4k$
Intensity Ratio:
$$ \frac{I_{max}}{I_{min}} = \frac{196k}{4k} = \frac{49}{1} $$
Answer:
The ratio of maximum to minimum intensities produced by the superposition of these waves is $49:1$.
Correct option: A) $49:1$.
Intensity level $2 \mathrm{~cm}$ from a source of sound is $80 \mathrm{~dB}$. If there is no loss of acoustic power in air and intensity of threshold hearing is $10^{-12} \mathrm{~W} \mathrm{~m}^{-2}$, what is the intensity level at a distance of $40 \mathrm{~cm}$ from the source?
A) Zero B) $54 \mathrm{~dB}$ C) $64 \mathrm{~dB}$ D) $44 \mathrm{~dB}$
The correct answer is B) 54 dB.
Here, the intensity of sound $I$ at a location from a point source is inversely proportional to the square of the distance from the source. Mathematically, this relation can be expressed as:
$$ I \propto \frac{1}{r^2} $$
where $r$ is the distance from the source. Thus, the ratio of intensities at two different distances $r_1$ and $r_2$ is given by:
$$ \frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} = \frac{2^2}{40^2} = \frac{1}{400} $$
$I_1$ and $I_2$ represent the intensities at $2$ cm and $40$ cm, respectively.
Using the logarithmic scale for intensity levels (in decibels), the sound intensity level (SIL) can be calculated as:
$$ SIL = 10 \log_{10} \left( \frac{I}{I_0} \right) $$
where $I_0 = 10^{-12} \mathrm{W/m^2}$ is the threshold intensity of hearing.
As $I_1 = 400 I_2$, their corresponding intensity levels are:
$$ I_1 - I_2 = 10 \log_{10}\left( \frac{I_1}{I_2} \right) = 10 \log_{10}(400) $$
Calculating $\log_{10}(400)$ we get:
$$ \log_{10}(400) = 2.602 $$
Thus, the difference between their levels:
$$ I_1 - I_2 = 10 \times 2.602 = 26 \text{ dB} $$
Given that the intensity level at 2 cm is $80 \text{ dB}$, at 40 cm it would be:
$$ I_2 = I_1 - 26 = 80 - 26 = 54 \text{ dB} $$
Therefore, the intensity level at a distance of 40 cm from the source is 54 dB.
A $20 \mathrm{~cm}$ long string, having a mass of $1.0 \mathrm{~g}$, is fixed at both ends. The tension in the string is $0.5 \mathrm{~N}$. The string is set into vibrations using an external vibrator of frequency $100 \mathrm{~Hz}$. Find the separation (in $\mathrm{cm}$) between the successive nodes on the string.
To find the separation between successive nodes on the vibrating string, we'll follow these steps:
Calculate the mass per unit length (linear density) of the string: $$ \mu = \frac{1.0 \times 10^{-3} , \text{kg}}{20 \times 10^{-2} , \text{m}} = 5 \times 10^{-3} , \text{kg/m} $$
Determine the speed of waves on the string using the formula for wave speed in a string under tension: $$ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{0.5}{5 \times 10^{-3}}} = 10 , \text{m/s} $$
Calculate the wavelength using the relation between wave speed, frequency, and wavelength: $$ v = f \lambda \quad \Rightarrow \quad \lambda = \frac{v}{f} = \frac{10}{100} = 0.1 , \text{m} = 10 , \text{cm} $$
Find the separation between successive nodes, which is half the wavelength: $$ \text{Separation} = \frac{\lambda}{2} = \frac{10 , \text{cm}}{2} = 5 , \text{cm} $$
Therefore, the separation between the successive nodes on the string is 5 cm.
A longitudinal wave is propagating along the $x$-axis. Particles of the medium vibrate along
A $x$-axis
B $y$-axis
C $z$-axis
D None of these
The correct option is A) $x$-axis.
In a longitudinal wave, particles of the medium vibrate along the direction of the wave's propagation. Since the wave is propagating along the $x$-axis, the vibrations of the particles also occur along the $x$-axis.
Therefore, option A is the correct answer.
Which two seas are connected by the Kiel Canal?
A. North Sea and Baltic Sea
B. Mediterranean and Black Sea
C. Pacific and Atlantis Oceans
D. Mediterranean and Red Sea
The correct answer is A. North Sea and Baltic Sea.
The Kiel Canal serves as a crucial waterway, connecting the North Sea and the Baltic Sea.
Why are microwaves suitable for radar systems used in aircraft navigation?
Microwaves are a preferred choice for radar systems in aircraft navigation due to their shorter wavelengths, which provide high resolution for target detection. This allows microwaves to effectively bounce off smaller objects, making it easier to detect even smaller aircraft. The utility of different wavelengths allows radars to be specialized for various functions:
Weather radars and surface search radars operate at distinct frequencies suited to their specific tasks.
Target acquisition and target tracking radars also use different frequencies, each optimized for their respective roles.
Missile guidance systems use yet another frequency tailored to their operational requirements.
The choice of frequency is crucial because it affects the radar's performance in real-world scenarios. For instance, radiowaves, with lower frequencies and longer wavelengths, can bend around the Earth's curvature better but might require more power and could miss smaller targets like stealth drones.
Moreover, microwaves have a significant advantage over infrared waves in terms of penetration capabilities. While infrared waves have much shorter wavelengths, microwaves can penetrate clouds and provide data from various altitudes, providing a more robust performance in varied weather conditions. In contrast, infrared systems have lesser penetration capabilities, limiting their effectiveness to lower altitudinal information.
"A wave transfers:
A. Matter
B. Energy
C. Liquid
D. Gas"
The correct answer is B. Energy.
Waves are primarily a method for the transfer of energy from one point to another without the transportation of matter. Unlike options A, C, and D, which suggest the movement of matter or specific states of matter (liquid, gas), a wave itself does not carry matter, but rather energy through the medium. Whether the medium is solid, liquid, or gas, it's the energy that is transmitted, not the material substance. Therefore, the option B. Energy is the apt choice.
When water rises to its highest level, covering much of the seashore, we call it:
A) a high tide
B) a large tidal wave
C) a low tide
D) a big tide
The correct answer is A) a high tide.
Tides are the periodic rise and fall of ocean water, occurring typically twice a day. A high tide happens when ocean water reaches its maximum height, covering much of the seashore. In contrast, a low tide represents times when the ocean's water pulls back, receding to its lowest levels along the shore.
The process of pushing the oceanic crust on the sides of ridges and the spreading of the ocean floor is known as $\qquad$.
A Tides
B Seafloor mapping
C Seafloor spreading
D Sea waves
The correct answer is C) Seafloor spreading.
Seafloor spreading is the process that involves the pushing of the oceanic crust on the sides of ridges and the spreading of the ocean floor. This phenomenon contributed significantly to refining and questioning existing geological theories, including Wegener's Theory.
Select the most suitable options that represent the movements of the ocean water.
A) leaping
B) currents
C) waves
D) tides
The most appropriate terms describing the movements of ocean water are:
B) currents
C) waves
D) tides
Oceanic movements are broadly classified into waves, currents, and tides:
Waves occur when the water on the ocean's surface rises and falls.
Tides refer to the periodic movements of ocean water, influenced primarily by the gravitational pull of the Moon and the Sun.
Ocean currents represent channels of water moving in a specific direction within the oceans.
The displacement due to a wave moving in the positive $x$-direction is given by $y = \frac{1}{\left(1 + x^{2}\right)}$ at time $t = 0$ and by $y = \frac{1}{1 + (x-1)^{2}}$ at $t = 2$ seconds, where $x$ and $y$ are in meters. The velocity of the wave in m/s is:
A) 0.5
B) 1
C) 2
D) 4
The correct answer is Option A: $0.5$ m/s.
The displacement due to the wave can be written in a general form that includes both position $x$ and time $t$. For a wave traveling in the positive $x$-direction, we can use the form: $$ y = \frac{1}{1 + (x - vt)^2} $$ where $v$ is the velocity of the wave.
At time $t = 0$, we have the equation: $$ y = \frac{1}{1 + x^2} $$ which matches the given $y = \frac{1}{1 + x^2}$ at $t = 0$ when $v = 0$.
At time $t = 2$ seconds, the reference equation becomes: $$ y = \frac{1}{1 + (x - 2v)^2} $$ We are given $y = \frac{1}{1 + (x - 1)^2}$ at $t = 2$ seconds, which implies: $$ x - 2v = x - 1 $$ Solving for $v$, we find: $$ 2v = 1 $$ $$ v = 0.5 , \text{m/s} $$ Thus, the velocity of the wave is $0.5$ m/s.
A source $S$ emitting sound of 300 Hz is fixed on block A which is attached to the free end of spring $S_{A}$ as shown in the figure. The detector D fixed on block B attached to the free end of spring $S_{B}$ detects this sound.
The blocks A and B are simultaneously displaced towards each other through 1.0 m and then left to vibrate. Find the maximum and minimum frequencies of sound detected by D if the vibrational frequency of each block is 2 Hz (velocity of sound is 340 m/s).
A. 323 Hz, 278.6 Hz
B. 276.8 Hz, 330 Hz
C. 260 Hz , 241 Hz
D. 340 Hz , 260 Hz
To determine the maximum and minimum frequencies of the sound detected by the detector ( D ) when blocks ( A ) and ( B ) vibrate, follow the steps below:
Given:
Source frequency ( f ): 300 Hz
Velocity of sound ( v ): 340 m/s
Vibrational frequency of each block: 2 Hz
Displacement amplitude ( A ): 1 m
Calculation:
Angular frequency $ \omega $: [ \omega = 2 \pi f = 2 \pi \times 2 = 12.56 , \text{rad/s} ]
Maximum velocity $v_m$: [ v_m = A \omega = 1 \times 12.56 = 12.56 , \text{m/s} ]
Maximum Frequency $ f_{\text{max}}$
When both blocks move towards each other: [ f_{\text{max}} = \left( \frac{v + v_d}{v - v_s} \right) f = \left( \frac{340 + 12.56}{340 - 12.56} \right) 300 ] [ f_{\text{max}} = \left( \frac{352.56}{327.44} \right) 300 = 1.0768 \times 300 = 323 , \text{Hz} ]
Minimum Frequency $f_{\text{min}} $
When both blocks move away from each other: [ f_{\text{min}} = \left( \frac{v - v_d}{v + v_s} \right) f = \left( \frac{340 - 12.56}{340 + 12.56} \right) 300 ] [ f_{\text{min}} = \left( \frac{327.44}{352.56} \right) 300 = 0.9282 \times 300 = 278.6 , \text{Hz} ]
Final Answer:
Maximum frequency: 323 Hz
Minimum frequency: 278.6 Hz
A man of mass 50 kg is running on a plank of mass 150 kg with a speed of 8 m/s relative to the plank as shown in the figure (both were initially at rest and velocity of man with respect to ground anyhow remains constant). The plank is placed on a smooth horizontal surface. The man, while running, whistles with frequency $f_{0}$. A detector (D) placed on the plank detects the frequency. The man jumps off with the same velocity (with respect to the ground) from point D and slides on the smooth horizontal surface (Assume the coefficient of friction between man and horizontal is zero). The speed of sound in a still medium is 330 m/s.
The frequency of sound detected by detector D before the man jumps off the plank is:
A) $\frac{332}{324} f_{0}$
B) $\frac{330}{322} f_{0}$
C) $\frac{328}{336} f_{0}$
D) $\frac{328}{338} f_{0}$
Given the problem, we need to find the frequency detected by detector D before the man jumps off the plank. The scenario includes a man of mass 50 kg running on a plank with mass 150 kg with a speed of 8 m/s relative to the plank. The plank is placed on a smooth horizontal surface.
Let's denote the various parameters we have:
Mass of the man, $m_1 = 50$ kg
Mass of the plank, $m_2 = 150$ kg
Velocity of the man with respect to the plank, $v_{\text{man, plank}} = 8$ m/s
Speed of sound in air, $v_{\text{sound}} = 330$ m/s
Since the surface is frictionless and no external horizontal force acts on the system, we use the conservation of linear momentum:
[ m_1 v_{\text{man, ground}} = m_2 v_{\text{plank, ground}} ]
Step 1: Find the Velocities
Assume the velocity of the plank with respect to the ground is $v_{\text{plank, ground}} = v$.
The velocity of the man with respect to the ground is $v_{\text{man, ground}} = 8 - v$.
Applying conservation of momentum:
[ m_1 \cdot v_{\text{man, ground}} = m_2 \cdot v_{\text{plank, ground}} ]
[ 50 \cdot (8 - v) = 150 \cdot v ]
Solving for $v$:
[ 50 \cdot 8 - 50v = 150v ]
[ 400 = 200v ]
[ v = \frac{400}{200} = 2 , \text{m/s} ]
Thus: [ v_{\text{plank, ground}} = 2 , \text{m/s} ]
[ v_{\text{man, ground}} = 8 - 2 = 6 , \text{m/s} ]
Step 2: Apply the Doppler Effect
The frequency detected by the detector can be calculated using the Doppler Effect formula:
[ f' = f_{0} \left( \frac{v_{\text{sound}} + v_{\text{observer}}}{v_{\text{sound}} - v_{\text{source}}} \right) ]
Here:
The observer (detector D) moves with the plank at $v_{\text{plank, ground}} = 2 , \text{m/s}$.
The source (man) moves at $v_{\text{man, ground}} = 6 , \text{m/s}$ towards the detector.
Plugging these into the Doppler Effect equation:
[ f' = f_{0} \left( \frac{330 + 2}{330 - 6} \right) ]
[ f' = f_{0} \left( \frac{332}{324} \right) ]
The detected frequency before the man jumps off the plank is:
[ \boxed{\frac{332}{324} f_{0}} ]
Thus, the correct answer is (A) $\frac{332}{324} f_0$.
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Ask Chatterbot AINCERT Solutions - Waves | NCERT | Physics | Class 11
A string of mass $2.50 \mathrm{~kg}$ is under a tension of $200 \mathrm{~N}$. The length of the stretched string is $20.0 \mathrm{~m}$. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
The time it takes for the disturbance to travel the length of the string is 0.5 seconds.
Hence, the disturbance takes 0.5 seconds to reach the other end of the string.
A stone dropped from the top of a tower of height $300 \mathrm{~m}$ 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 $340 \mathrm{~m} \mathrm{~s}^{-1}$ ? $\left(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)$
The total time to hear the splash at the top of the tower involves two steps:
Time for the stone to fall $300 \mathrm{~m}$ with $g=9.8 \mathrm{~m} \mathrm{~s}^{-2}$: 7.8246 seconds
Time for the sound to travel back up $300 \mathrm{~m}$ at $340 \mathrm{~m} \mathrm{~s}^{-1}$: 0.8824 seconds
The total time is the sum of these two durations:
$$ t_{\text{total}} = 7.8246 + 0.8824 = 8.707 \text{ seconds} $$
The splash is heard at the top 8.707 seconds after the stone is dropped.
A steel wire has a length of $12.0 \mathrm{~m}$ and a mass of $2.10 \mathrm{~kg}$. 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 $20^{\circ} \mathrm{C}=343 \mathrm{~m} \mathrm{~s}^{-1}$.
The tension ( T ) required in the steel wire so that the speed of a transverse wave equals$ 343 \mathrm{~m} \mathrm{~s}^{-1}$ is:
[ \mathbf{T = 20588.575 , \mathrm{N}} ]
So, you need to apply a tension of approximately 20588.6 N.
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.
Let's break down the formula for the speed of sound in air to address the three points:
Formula for the Speed of Sound in Air
$$ v = \sqrt{\frac{\gamma P}{\rho}} $$
Where:
$ v $ is the speed of sound
$ \gamma $ is the adiabatic index (ratio of specific heats)
$ P $ is the pressure of the air
$ \rho $ is the density of the air
(a) Independence of Speed of Sound from Pressure
The formula can be rewritten using the ideal gas law $P = \rho R T$, where $R$ is the specific gas constant for air, and $T$ is the temperature in Kelvin. Substituting $\rho = \frac{P}{R T}$ into the speed of sound equation, we get:
$$ v = \sqrt{\frac{\gamma P}{\frac{P}{R T}}} = \sqrt{\frac{\gamma R T}{1}} = \sqrt{\gamma R T} $$
Explanation:
The pressure $P$ cancels out in the equation $v = \sqrt{\gamma R T}$.
Therefore, the speed of sound in air is independent of pressure.
(b) Increase of Speed of Sound with Temperature
Using the relation derived above: $$ v = \sqrt{\gamma R T} $$
Explanation:
The speed of sound $v$ is proportional to the square root of the absolute temperature $T$.
As the temperature $T$ increases, the speed of sound $v$ increases because $\sqrt{\gamma R T}$ increases.
(c) Increase of Speed of Sound with Humidity
Humidity affects the density $\rho$ of air:
When air is humid, water vapor (which has a lower molar mass) replaces some of the nitrogen and oxygen molecules (which have higher molar masses).
This leads to a decrease in the density $\rho$ of the air because water vapor is less dense than dry air.
Explanation:
From the equation $v = \sqrt{\frac{\gamma P}{\rho}}$, if $\rho$ decreases while $\gamma$ and $P$ remain constant, the speed of sound $v$ increases.
Thus, the speed of sound increases with humidity because the density of humid air is lower compared to dry air.
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. $y=f(x \pm v t)$. Is the converse true? Examine if the following functions for $y$ can possibly represent a travelling wave :
(a) $(x-v t)^{2}$
(b) $\log \left[(x+v t) / x_{0}\right]$
(c) $1 /(x+v t)$
1. $y = (x - v t)^2 $
This function is of the form $ (x - v t)^2$. While $x - v t $ represents the combination typically seen in a travelling wave, the square of this term results in a quadratic form which doesn't match the standard form of a travelling wave function $ f (x \pm v t) $. Therefore, this function does not represent a travelling wave.
2. $y = \log \left[ \frac{x + v t}{x_0} \right] $
This function is of the form $ \log \left( \frac{x + v t}{x_0} \right) $. Here, $x + v t $ represents the combination in a logarithmic form. Although $ x + v t $ is seen in the combination within the function, the logarithmic nature does not conform to a simple sinusoidal or travelling wave form. Therefore, this function does not represent a travelling wave.
3. ( y = \frac{1}{x + v t} )
This function is of the form $\frac{1}{x + v t} $. It contains the combined terms $ x + v t $ in the denominator. Similar to the previous evaluations, although $x + v t $ is in the required form, the reciprocal nature deviates from the standard sinusoidal or travelling wave form. Therefore, this function does not represent a travelling wave.
Summary:
(a) $ (x - v t)^2 $: Does not represent a travelling wave.
(b) $\log \left[ \frac{x + v t}{x_0} \right] $: Does not represent a travelling wave.
(c) $\frac{1}{x + v t} $: Does not represent a travelling wave.
A bat emits ultrasonic sound of frequency $1000 \mathrm{kHz}$ 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 $340 \mathrm{~m} \mathrm{~s}^{-1}$ and in water $1486 \mathrm{~m} \mathrm{~s}^{-1}$.
There seems to be a misunderstanding with the inputs processed earlier since they were returned as acceleration instead of frequencies and wavelengths. Let's manually calculate the values for the correct results:
(a) Wavelength of the Reflected Sound
Given:
Speed of sound in air $( v_\text{air} )) = ( 340 \ \mathrm{m/s} $
Frequency ($ f $) = $ 1000 \times 10^3 \ \mathrm{Hz} $
[ \lambda_\text{air} = \frac{v_\text{air}}{f} = \frac{340 \ \mathrm{m/s}}{1000 \times 10^3 \ \mathrm{Hz}} = \frac{340}{10^6} \ \mathrm{m} = 0.34 \ \mathrm{mm} ]
(b) Wavelength of the Transmitted Sound
Given:
Speed of sound in water ($ v_\text{water} $) = $ 1486 \ \mathrm{m/s} $
Frequency ($ f $) = $ 1000 \times 10^3 \ \mathrm{Hz} $
[ \lambda_\text{water} = \frac{v_\text{water}}{f} = \frac{1486 \ \mathrm{m/s}}{1000 \times 10^3 \ \mathrm{Hz}} = \frac{1486}{10^6} \ \mathrm{m} = 1.486 \ \mathrm{mm} ]
Summary
Wavelength of the reflected sound in air: $0.34 \ \mathrm{mm}$
Wavelength of the transmitted sound in water: $1.486 \ \mathrm{mm}$
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 $1.7 \mathrm{~km} \mathrm{~s}^{-1}$ ? The operating frequency of the scanner is $4.2 \mathrm{MHz}$.
The wavelength $ \lambda $ of the sound in the tissue can be calculated as follows:
[ \lambda = \frac{1.7 \times 10^3 , \text{m/s}}{4.2 \times 10^6 , \text{Hz}} = 4.0476 \times 10^{-4} , \text{m} ]
So, the wavelength of the sound in the tissue is approximately 0.404 mm.
A transverse harmonic wave on a string is described by
$$ y(x, t)=3.0 \sin (36 \mathrm{t}+0.018 x+\pi / 4) $$
where $x$ and $y$ are in $\mathrm{cm}$ and $t$ in $\mathrm{s}$. 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?
(a) Type of Wave and Propagation
Wave Speed: [ v = \frac{\omega}{k} = \frac{36}{0.018} = 2000 \text{ cm/s} = 20 \text{ m/s} ]
Direction of Propagation: Negative ( x )-direction
(b) Amplitude and Frequency
Amplitude: ( 3.0 \text{ cm} )
Frequency: [ f = \frac{\omega}{2\pi} = \frac{36}{2\pi} \approx 5.73 \text{ Hz} ]
(c) Initial Phase at the Origin
Initial Phase: ( \frac{\pi}{4} )
(d) Distance Between Successive Crests
Wavelength: [ \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.018} \approx 349.07 \text{ cm} ]
[ \boxed{ \begin{aligned} & \text{(a) The wave is a travelling wave. It travels in the negative } x \text{-direction with a speed of } 20 \text{ m/s.} \ & \text{(b) The amplitude of the wave is } 3.0 \text{ cm and the frequency is } 5.73 \text{ Hz.} \ & \text{(c) The initial phase at the origin is } \frac{\pi}{4}. \ & \text{(d) The least distance between two successive crests is } 349.07 \text{ cm.} \end{aligned}} ]
These are the characteristics of the given wave.
For the wave described in Exercise 14.8, plot the displacement $(y)$ versus ($t$) graphs for $x=0,2$ and $4 \mathrm{~cm}$. 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?
Displacement versus Time Graphs
Here are the displacement $(y)$ vs. time $(t)$ graphs for different values of $x$:
For ( x = 0 ) cm:
For ( x = 2 ) cm:
For ( x = 4 ) cm:
Shapes and Differences
Shape of Graphs: All graphs are sinusoidal waves and have the same shape but are shifted along the time axis.
Differences in Oscillatory Motion:
Amplitude: The amplitude remains the same for all points (it does not change with $x$ in a traveling wave).
Frequency: The frequency also remains the same for all points (it does not depend on $x$).
Phase: The phase changes with different $x$ values. This is evident from the horizontal shifts in the sinusoidal graphs.
Thus, the phase is the aspect where the oscillatory motion in a travelling wave differs from one point to another.
For the travelling harmonic wave
$$ y(x, t)=2.0 \cos 2 \pi(10 t-0.0080 x+0.35) $$
where $x$ and $y$ are in $\mathrm{cm}$ and $t$ in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) $4 \mathrm{~m}$,
(b) $0.5 \mathrm{~m}$
(c) $\lambda / 2$,
(d) $3 \lambda / 4$
It seems like there was some confusion in interpreting the input properly. Let's simplify this. We already have $ k = 2 \pi \times 0.0080 = 0.05027 , \text{rad/cm} $. We'll manually calculate the phase differences using this value of ( k ).
(a) Separation $ \Delta x = 400 , \text{cm} $
[ \Delta \phi = k \Delta x = 0.05027 \times 400 , \text{rad} = 20.11 , \text{rad} ]
(b) Separation $ \Delta x = 50 , \text{cm} $
[ \Delta \phi = k \Delta x = 0.05027 \times 50 , \text{rad} = 2.51 , \text{rad} ]
Wavelength $ \lambda $
[ k = \frac{2 \pi}{\lambda} \implies \lambda = \frac{2 \pi}{k} = \frac{2 \pi}{0.05027} \approx 125 , \text{cm} ]
(c) Separation $ \Delta x = \lambda / 2 $
[ \Delta \phi = k \left( \frac{\lambda}{2} \right) = k \left( \frac{125}{2} \right) , \text{rad} = 0.05027 \times 62.5 , \text{rad} = 3.14 , \text{rad} ]
(d) Separation $\Delta x = 3\lambda / 4 $
[ \Delta \phi = k \left( \frac{3\lambda}{4} \right) = k \left( \frac{3 \times 125}{4} \right) , \text{rad} = 0.05027 \times 93.75 , \text{rad} = 4.71 , \text{rad} ]
Thus, the phase differences are:
(a) $ 4.0 , \text{m} $ : $ 20.11 , \text{rad} $
(b) $0.5 , \text{m} $ : $ 2.51 , \text{rad}$
(c) $ \lambda / 2 $ : $ 3.14 , \text{rad}$
(d) $ 3 \lambda / 4$ : $ 4.71 , \text{rad}$
The transverse displacement of a string (clamped at its both ends) is given by
$$ y(x, t)=0.06 \sin \frac{2 \pi}{3} x \cos (120 \pi t) $$
where $x$ and $y$ are in $\mathrm{m}$ and $t$ in $\mathrm{s}$. The length of the string is $1.5 \mathrm{~m}$ and its mass is $3.0 \times 10^{-2} \mathrm{~kg}$
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 ?
(c) Determine the tension in the string.
(a) Type of the Wave
The given wave function: $$ y(x, t) = 0.06 \sin \left( \frac{2 \pi}{3} x \right) \cos(120 \pi t) $$ represents a stationary wave.
(b) Superposition of Opposite Traveling Waves
Rewriting the stationary wave as a superposition of two traveling waves: $$ y_1 = \frac{0.06}{2} \sin \left( \frac{2 \pi}{3} x - 120 \pi t \right) $$ $$ y_2 = \frac{0.06}{2} \sin \left( \frac{2 \pi}{3} x + 120 \pi t \right) $$
Wavelength $\lambda$: $$ \frac{2\pi}{k} = \frac{2\pi}{\frac{2\pi}{3}} = 3 , m $$
Frequency $f$: $$ f = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60 , Hz $$
Speed $v$: $$ v = \lambda f = 3 , m \times 60 , Hz = 180 , m/s $$
(c) Tension in the String
Using the formula: $$ v = \sqrt{\frac{T}{\mu}} $$
Given the linear mass density $\mu$: $$ \mu = \frac{3.0 \times 10^{-2} , \text{kg}}{1.5 , \text{m}} = 0.02 , \text{kg/m} $$
The speed of the wave $v$: $$ v = 180 , \text{m/s} $$
We calculated the tension $T$: $$ T = \mu v^2 = 0.02 , \text{kg/m} \times (180 , \text{m/s})^2 = \boxed{648 , N} $$
Therefore, the tension in the string is 648 N.
(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 $0.375 \mathrm{~m}$ away from one end?
(i)For the wave on a string described in Exercise 15.11 (assuming it's a standing wave):
Frequency
(a) Do all the points on the string oscillate with the same frequency?
Yes, all points on the string oscillate with the same frequency. This is a characteristic of standing waves. The frequency of oscillation is determined by the source of the wave and is the same for all points on the string.
Phase
(b) Do all the points on the string oscillate with the same phase?
No, the points on the string do not all oscillate with the same phase. Points between nodes have different phases; their phase varies along the length of the string, except at the nodes where the displacement is always zero.
Amplitude
(c) Do all the points on the string oscillate with the same amplitude?
No, the amplitude is not the same for all points on the string. In a standing wave, the amplitude varies along the length of the string. It is zero at the nodes and maximum at the antinodes.
Amplitude of a point $0.375 \mathrm{~m}$ away from one end
(ii) To calculate the amplitude of a point $0.375 \mathrm{~m}$ away from one end, we need to use the equation of the standing wave. Generally, for a standing wave of amplitude $A$,
$$ y(x,t) = 2A \sin(kx) \cos(\omega t) $$
Here, ( x ) represents the position along the string.
The amplitude of a point ( x ) away from one end can be given by:
$$ A(x) = 2A \sin(kx) $$
Now, let’s calculate the amplitude at ( x = 0.375 \mathrm{~m} ):
Step 1: Find the wave number ( k )
Typically, ( k ) is given by:
$$ k = \frac{2\pi}{\lambda} $$
Step 2: Use the amplitude equation for ( x = 0.375 \mathrm{~m} )
Using the general wave equation,
$$ A(x) = 2A \sin(kx) $$
Inputting ( x = 0.375 \mathrm{~m} ),
$$ A(0.375) = 2A \sin\left( k \times 0.375 \right) $$
Since there's no information on the specific wave number ( k ) or the amplitude ( A ), we cannot provide a numerical result without this information. But this formula gives the relationship required.
Please provide the wave number ( k ) or wavelength ( \lambda ) and the amplitude ( A ) if available to get the exact amplitude value.
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) $y=2 \cos (3 x) \sin (10 t)$
(b) $y=2 \sqrt{x-v t}$
(c) $y=3 \sin (5 x-0.5 t)+4 \cos (5 x-0.5 t)$
(d) $y=\cos x \sin t+\cos 2 x \sin 2 t$
(a) $ y = 2 \cos (3 x) \sin (10 t) $
This function can be rewritten using the trigonometric identity:
[ y = 2 \cos (3 x) \sin (10 t) = 2 \left[ \cos (3 x) \right] \left[ \sin (10 t) \right] ]
In this form, it represents the product of a function of (x) and a function of (t), indicating that it is a stationary wave.
(b) $ y = 2 \sqrt{x - v t}$
This function has the form:
[ y = f(x - v t) ]
It depends on ((x - v t)), which is characteristic of a travelling wave.
(c) $ y = 3 \sin (5 x - 0.5 t) + 4 \cos (5 x - 0.5 t) $
This function consists of two components that both depend on ($5 x - 0.5 t$). This can be simplified into a single wave function through trigonometric identities:
[ y = A \sin (5 x - 0.5 t + \phi) ]
where (A) and (\phi) are constants. Since it can be represented in the form (g(k x - \omega t)), it is a travelling wave.
(d) $ y = \cos x \sin t + \cos 2 x \sin 2 t $
This function does not have a form like $f(x - v t)$ or $g(x + v t)$ nor does it match the product of separate functions of (x) and (t). Therefore, it represents none at all.
Summarizing:
(a) $ y = 2 \cos (3 x) \sin (10 t) $: Stationary wave
(b) $y = 2 \sqrt{x - v t}$: Travelling wave
(c) $y = 3 \sin (5 x - 0.5 t) + 4 \cos (5 x - 0.5 t) $: Travelling wave
(d) $ y = \cos x \sin t + \cos 2 x \sin 2 t $: None at all
These classifications follow from the properties and forms of wave functions described in wave mechanics.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45 \mathrm{~Hz}$. The mass of the wire is $3.5 \times 10^{-2} \mathrm{~kg}$ and its linear mass density is $4.0 \times 10^{-2} \mathrm{~kg} \mathrm{~m}^{-1}$. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
(a) Speed of the Transverse Wave
Length ( L ) of the wire:
$$ L = \frac{m}{\mu} = 0.875 \text{ meters} $$
Speed ( v ) of the transverse wave:
Using ( v = 2Lf ):
$$ v = 78.75 \text{ meters/second} $$
(b) Tension in the String
Using the formula ( T = \mu v^2 ):
$$ T = 248.063 \text{ N} $$
Final Results
Speed of the transverse wave ( v ):[ v = 78.75 , \text{meters/second} ]
Tension in the string ( T ):[ T = 248.063 , \text{newtons} ]
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 $340 \mathrm{~Hz}$ ) when the tube length is $25.5 \mathrm{~cm}$ or $79.3 \mathrm{~cm}$. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Harmonic Lengths Analysis
$ L_1 = 25.5 , \text{cm} = 0.255 , \text{m} $
$ L_2 = 79.3 , \text{cm} = 0.793 , \text{m} $
Fundamental Frequency $ f = 340 , \text{Hz} $
For a pipe open at one end and closed at the other (with odd harmonics): [ \text{Length } (L) = (n + \frac{1}{2}) \frac{\lambda}{2} ]
Based on the provided lengths:
First Resonance
[ L_1 = \frac{\lambda}{4} ] [ 0.255 , \text{m} = \frac{\lambda}{4} \implies \lambda = 1.02 , \text{m} ]
Third Resonance
[ L_2 = \frac{3\lambda}{4} ] [ 0.793 , \text{m} = \frac{3\lambda}{4} \implies \lambda = 1.06 , \text{m} \approx 1.02 , \text{m} ]
Consistently, wavelength $\lambda \approx 1.02 , \text{m}$.
Speed of Sound Calculation
Using the resonance condition: [ v = f \lambda ] [ v = 340 , \text{Hz} \times 1.02 , \text{m} \approx 346.8 , \text{m/s} ]
Thus, the estimated speed of sound in air at the temperature of the experiment is: [ \boxed{346.8 , \text{m/s}} ]
A steel rod $100 \mathrm{~cm}$ long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53 \mathrm{kHz}$. What is the speed of sound in steel?
The speed of sound in steel is:
$$ v = 2 \times 1 , \text{m} \times 2530 , \text{Hz} = 5060 , \text{m/s} $$
So, the speed of sound in steel is 5060 m/s.
A pipe $20 \mathrm{~cm}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430 \mathrm{~Hz}$ source ? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is $340 \mathrm{~m} \mathrm{~s}^{-1}$ ).
For the pipe closed at one end:
Solving the equation [ \left( n + 0.5 \right) \frac{340}{4 \times 0.2} = 430 ]
[ 425 \left( n + 0.5 \right) = 430 ]
[ n + 0.5 \approx 1.011 ]
[ n \approx 0.511 ]
Since ( n ) must be an integer, the closest value is ( n = 1 ). Therefore, the first harmonic mode (( n = 1 )) of the pipe closed at one end resonates with the $ 430 , \text{Hz} $ source.
For the pipe open at both ends:
Solving the equation [ n \frac{340}{2 \times 0.2} = 430 ]
[ 850 n = 430 ]
[ n \approx 0.506 ]
Since ( n ) must be an integer for resonance, but here $ n \approx 0.506 $ means there is no integer ( n ) that satisfies this condition. Therefore, the same $ 430 , \text{Hz} $ source will not resonate with the pipe if both ends are open.
Two sitar strings A and B playing the note ' Ga' are slightly out of tune and produce beats of frequency $6 \mathrm{~Hz}$. The tension in the string A is slightly reduced and the beat frequency is found to reduce to $3 \mathrm{~Hz}$. If the original frequency of A is $324 \mathrm{~Hz}$, what is the frequency of B?
Initial Beat Frequency: The original frequency of string A ($ f_A $) is 324 Hz. The beat frequency with string B ($ f_B $) is given as 6 Hz. This implies: [ | f_A - f_B | = 6 \text{ Hz} ] Therefore, the possible frequencies for $ f_B $ are: [ f_B = 324 \text{ Hz} + 6 \text{ Hz} = 330 \text{ Hz} \quad \text{or} \quad f_B = 324 \text{ Hz} - 6 \text{ Hz} = 318 \text{ Hz} ]
Changing the Tension: When the tension in string A is reduced, its frequency decreases. The new beat frequency becomes 3 Hz. Let's denote the new frequency of A as $ f_A' $.
Since the beat frequency is reduced, consider: [ | f_A' - f_B | = 3 \text{ Hz} ]
If $ f_B = 330 \text{ Hz} $, then: [ f_A' = 330 \text{ Hz} \pm 3 \text{ Hz} \rightarrow f_A' = 333 \text{ Hz} \quad \text{or} \quad f_A' = 327 \text{ Hz} ]
Both cases result in an increase in frequency, which contradicts the fact that reducing tension decreases frequency.
Therefore, $ f_B $ cannot be 330 Hz.
Conclusion: Hence, $ f_B$ must be 318 Hz. Let's verify: [ 324 \text{ Hz} - 318 \text{ Hz} = 6 \text{ Hz} ]
With reduced tension, [ f_A' = 318 \text{ Hz} \pm 3 \text{ Hz} \rightarrow f_A' = 321 \text{ Hz} \quad \text{or} \quad f_A' = 315 \text{ Hz} ]
Both cases result in a decrease in frequency, which is consistent with reducing tension.
Therefore, the frequency of string B ($ f_B $) is 318 Hz.
Explain why (or how):
(a) in a sound wave, a displacement node is a pressure antinode and vice versa,
(b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any "eyes",
(c) a 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.
(a) In a sound wave, a displacement node is a pressure antinode and vice versa
In a sound wave propagating through a medium like air, the displacement of particles and pressure variations are closely related but occur in different phases:
Displacement Node: At these points, the particles of the medium have zero displacement; they do not move back and forth. Since particles are not moving, it implies a high density or pressure because the adjacent particles are maximally compressed. Thus, a displacement node corresponds to a pressure antinode where pressure variation is maximum.
Displacement Antinode: At these points, particles have maximum displacement. These regions correspond to points where the pressure variation is minimal (rarefaction). Hence, a displacement antinode is a pressure node where pressure variation is minimum.
(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any "eyes"
Bats use a method called echolocation to "see" in the dark:
Emission of Ultrasonic Waves: Bats emit high-frequency sound waves (ultrasonic waves) from their mouth or nose.
Reflection and Echo Detection: These sound waves hit objects and reflect back as echoes.
Analysis of Echoes: By analyzing the time delay between the emission of the sound wave and the returning echo, bats can determine the distance to the object. They can also infer the size, shape, and texture of the object based on the strength and characteristics of the echo.
Direction Determination: The direction of the obstacle can be ascertained by detecting the direction from which the echo returns using the difference in arrival times of the echo at the two ears.
(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes
Although a violin and a sitar may produce notes of the same fundamental frequency, they are distinguishable because of timbre or tone quality:
Harmonics and Overtones: The specific combination and relative strength of harmonics and overtones produced by each instrument give it a unique sound.
Attack and Decay: The way a sound's amplitude increases at the beginning (attack) and diminishes at the end (decay) vary between instruments.
Construction and Material: The materials and construction of the instruments affect how they vibrate and thus the resultant sound produced.
(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases
Solids: In solids, both longitudinal (compression and expansion) and transverse (shear) waves can propagate because solids have both bulk modulus and shear modulus. Solids resist changes in shape, allowing them to support shear waves (transverse waves).
Gases: Gases can only support longitudinal waves because they lack shear strength (shear modulus). Gases cannot sustain shear stress, so only compressive and expansive deformations (longitudinal waves) can propagate.
(e) The shape of a pulse gets distorted during propagation in a dispersive medium
In a dispersive medium, the velocity of the wave depends on its frequency:
Dispersion: Different frequency components of a pulse travel at different speeds. As a result, different parts of the pulse (different frequencies) arrive at different times.
Pulse Distortion: This variation in speed causes the pulse to spread out and change shape as it propagates through the medium. Components of different frequencies interfere differently over time, leading to a distortion of the initial pulse shape.
These detailed explanations showcase fundamental concepts in wave mechanics and their applications in real-world scenarios.
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Notes - Waves | Class 11 NCERT | Physics
Comprehensive Class 11 Notes on Waves: Understanding the Fundamentals
Waves form a foundational concept in physics, with applications across various fields like acoustics, optics, and quantum mechanics. This guide serves as a comprehensive reference for Class 11 students to understand the fundamentals of waves.
Introduction to Waves
Waves are disturbances that transfer energy through a medium without the actual movement of matter. They are classified broadly into mechanical, electromagnetic, and matter waves.
- Mechanical Waves: These require a medium (solid, liquid, or gas) for their propagation.
- Electromagnetic Waves: These can travel through a vacuum and do not require a medium.
- Matter Waves: Associated with particles in quantum mechanics, like electrons and protons.
Types of Mechanical Waves
Transverse Waves
In transverse waves, particles of the medium move perpendicular to the direction of wave propagation. A common example is waves on a string.
Longitudinal Waves
In longitudinal waves, particles of the medium move parallel to the direction of wave propagation, such as sound waves in air.
Wave Propagation and Speed
Wave speed is determined by the medium's properties, like tension and mass density for a string, or bulk modulus and density for fluids.
-
For Transverse Waves: ( v = \sqrt{\frac{T}{\mu}} )
- Where ( T ) is tension and ( \mu ) is linear mass density.
-
For Longitudinal Waves (Sound): ( v = \sqrt{\frac{B}{\rho}} )
- Where ( B ) is bulk modulus and ( \rho ) is density.
Energy Transportation through Waves
Waves transport energy from one point to another. For example, sound waves carry energy through compressions and rarefactions in air, while electromagnetic waves (like light) carry energy through oscillating electric and magnetic fields.
Superposition Principle and Interference
The principle of superposition states that the resultant displacement at a point is the sum of the displacements due to individual waves.
Constructive and Destructive Interference
- Constructive Interference: When waves combine to produce a larger amplitude.
- Destructive Interference: When waves combine to cancel each other out.
Beats
Beats occur when two waves of slightly different frequencies superpose, resulting in a periodic increase and decrease in amplitude.
Reflection and Refraction of Waves
When waves encounter boundaries, they can be:
- Reflected: Echoes are a practical example.
- Refracted: Changing speed and direction when entering a different medium.
Phase Changes
- Rigid Boundary: Reflects with a phase change of ( 180^\circ ).
- Open Boundary: Reflects without a phase change.
Standing Waves and Resonance
Standing waves form when two waves of the same frequency and amplitude travel in opposite directions, leading to fixed points of no displacement called nodes and maximum displacement called antinodes.
graph LR
A(Start Point) -- Wave Propagation --> B(Standing Wave Formation)
B --> C(Nodes)
B --> D(Antinodes)
Harmonics
- Fundamental Frequency: The lowest frequency at which the system vibrates.
- Overtones: Higher harmonics.
Practical Applications and Examples
Waves on Strings
Musical instruments like guitars and violins produce sound through vibrating strings.
Sound Waves in Air and Solids
Sound travels faster in solids compared to liquids and gases due to higher density and elasticity.
Important Formulas and Equations
-
Wave Speed: ( v = \frac{\omega}{k} = \lambda \nu )
- Where ( \omega ) is angular frequency, ( k ) is wave number, ( \lambda ) is wavelength, and ( \nu ) is frequency.
- Displacement in a Travelling Wave: ( y(x, t) = a \sin(kx - \omega t + \phi) )
Conclusion
Understanding waves is crucial for grasping concepts in physics. They not only describe phenomena in everyday life but also in advanced fields like quantum physics. Ensuring a solid foundation in wave dynamics allows for a deeper appreciation of the natural world and technological applications.
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