Class 11 Physics (Physics Part-II) Chapter 14 Waves

Key Concept: Wave Combination

c) Transverse and longitudinal

[Solution Description] Ocean waves are comprised of both transverse and longitudinal wave components, resulting in their characteristic motion.

Key Concept: Amplitude Variation

b) When $\cos \omega_b t = +1$ or $-1$
[Solution Description] The amplitude of the resultant wave in a beat phenomenon is maximum when $\cos \omega_b t = +1$ or $-1$.

Key Concept: Wave Speed

b) 3 m

[Solution Description]
To find the wavelength $\lambda$, we use the relationship between speed $v$, frequency $\nu$, and wavelength $\lambda$:
$v = \lambda \nu$
Given that $v = 150$ m/s and $\nu = 50$ Hz, we can solve for $\lambda$:
$\lambda = \frac{v}{\nu} = \frac{150\ \text{m/s}}{50\ \text{Hz}} = 3\ \text{m}$
Thus, the wavelength is 3 meters.

Key Concept: Medium Requirement

b) Sound waves

[Solution Description] Mechanical waves require a medium to propagate, and sound is an example of such a wave.

Key Concept: Harmonic Identification

d) All of these

[Solution Description] In a pipe open at both ends, all harmonics (odd and even) are possible because there is an antinode at each end. Therefore, the first harmonic is also possible.

Key Concept: Comparative Analysis

c) Matter waves represent probabilities and are more abstract

[Solution Description] Unlike electromagnetic waves, which can propagate through a vacuum, matter waves are abstract and describe the probabilistic nature of particles in quantum mechanics.

Key Concept: Positive Amplitude

b) 6 cm

[Solution Description]

The amplitude ($a$) of a wave is the maximum positive displacement from the equilibrium position. Given that the displacement varies between -6 cm and 6 cm, the amplitude is:

$a = 6 \text{ cm}$

So, the amplitude is 6 cm.

Key Concept: Problem Solving

b) 600 Hz

[Solution Description]
The formula to calculate the observed frequency when the source is moving away is $f’ = \frac{f(v – v_o)}{(v + v_s)}$, where $f = 650\, \text{Hz}$, $v = 343\, \text{m/s}$, $v_o = 0\, \text{m/s}$, and $v_s = 30\, \text{m/s}$.
Substituting the values into the equation gives:
$f’ = \frac{650 \times (343 – 0)}{343 + 30}$
$f’ = \frac{650 \times 343}{373}$
$f’ = \frac{223450}{373}$
$f’ \approx 599.87\, \text{Hz}$
Thus, the frequency heard by the observer is approximately 600 Hz.

Key Concept: Wave Speed Factors

b) 1414.21 m/s

[Solution Description]

The speed of sound in a fluid is calculated using the formula $v = \sqrt{\frac{B}{\rho}}$. Substituting the given values:

$v = \sqrt{\frac{2 \times 10^9 \, \text{N/m}^2}{1000 \, \text{kg/m}^3}} = \sqrt{2 \times 10^6} = 1414.21 \, \text{m/s}$

Key Concept: Constructive Interference

b) $a_1 + a_2$
[Solution Description] Using the principle of superposition for constructive interference, when two waves are in phase, their amplitudes add up. Thus, the amplitude of the resultant wave is $a_1 + a_2$.

Key Concept: Formula Application

b) 0.628 s

[Solution Description]
We can find the period ($T$) using the formula $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{10} = \frac{\pi}{5} \approx 0.628 \, \text{s}$

Key Concept: Real-world Example

b) Transverse wave

[Solution Description] Dropping a pebble in water creates ripples that move outward from the point of impact. These ripples are examples of transverse waves because the water surface oscillates perpendicular to the direction of wave propagation.

Key Concept: Boundary Condition

a) Nodes at both ends

[Solution Description] A string fixed at both ends has nodes at each end due to zero displacement at these boundaries. Hence, the boundary condition is such that nodes are present at both fixed ends.

Key Concept: Formula Application

c) 1.2566 m

[Solution Description]
To find the wavelength ($λ$), use the formula for angular wave number:
$k = \frac{2π}{λ}.$
Rearrange to solve for $λ$:
$λ = \frac{2π}{k}.$
Substitute $k = 5$ rad/m:
$λ = \frac{2π}{5} = \frac{6.283}{5} = 1.2566 \, \text{m}.$

Key Concept: Wavelength Calculation

c) 4 m

[Solution Description]
The speed of a wave is given by the formula $v = \frac{\omega}{k}$, where $k$ is the wave number. To find the wavelength $\lambda$, use the relationship $k = \frac{2\pi}{\lambda}$. Therefore, rearrange the formula for $v$:
$v = \frac{\omega}{k} \implies k = \frac{\omega}{v}$
Since $k = \frac{2\pi}{\lambda}$, equate the two expressions and solve for $\lambda$:
$\frac{2\pi}{\lambda} = \frac{\omega}{v} \implies \lambda = \frac{2\pi v}{\omega}$
Substituting the given values,
$\lambda = \frac{2\pi \times 20}{10 \pi} = 4 \text{ m}$

Key Concept: Medium Requirement

d) Sound wave

[Solution Description] Among the given options, sound waves are mechanical waves that require a medium (like air, water, or solids) to propagate. Electromagnetic waves, on the other hand, do not need a medium.

Key Concept: Amplitude Calculation

c) $0$
[Solution Description] When two waves are out of phase by $\pi$, it results in destructive interference. In this case, the amplitude of the resultant wave is given by the formula $A(\phi) = 2a \cos(\phi/2)$. Substituting $\phi = \pi$, we have:
$A(\pi) = 2a \cos\left(\frac{\pi}{2}\right)$
Since $\cos(\pi/2) = 0$, the resultant amplitude is:
$A(\pi) = 2a \times 0 = 0$
Therefore, the amplitude of the resultant wave is zero.

Key Concept: Matter Waves and Applications

c) Electron microscopes

[Solution Description] Electron microscopes utilize the wave-like properties of electrons to achieve high-resolution imaging, taking advantage of their shorter wavelengths compared to visible light. This allows them to resolve structures at a much smaller scale.

Key Concept: Wave Characteristics

b) Longitudinal waves have parallel particle motion; transverse waves have perpendicular particle motion

[Solution Description] In transverse waves, particles move perpendicular to the direction of wave travel, whereas in longitudinal waves, particles move parallel to the direction of wave travel. This fundamental difference in motion distinguishes these two types of waves.

Key Concept: Frequency Adjustment

c) 5 Hz

[Solution Description]

The beat frequency ($\nu_{\text{beat}}$) can be found using the formula:

$\nu_{\text{beat}} = |\nu_1 – \nu_2|$

Substitute $\nu_1 = 53$ Hz and $\nu_2 = 48$ Hz into the equation:

$\nu_{\text{beat}} = |53 – 48| = 5 \text{ Hz}$

Key Concept: Definition Recall

a) $v = \sqrt{\frac{T}{\mu}}$

[Solution Description] The formula for the speed $v$ of a transverse wave on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension in the string and $\mu$ is the linear mass density.

Key Concept: Wave Equation Components

c) Initial phase angle

[Solution Description]
The term $\phi$ in the wave equation $y(x,t) = a \sin(kx – \omega t + \phi)$ represents the initial phase angle. It is the phase of the wave at position $x = 0$ and time $t = 0$. The initial phase angle determines the starting point of the wave oscillation.

Key Concept: Wave Equation

b) $y(x, t) = 0$
[Solution Description] Complete destructive interference occurs when the waves are completely out of phase such that their net displacement is zero everywhere and at all times. Given $y(x, t) = a \sin(kx – \omega t) + a \sin(kx – \omega t + \pi)$, using trigonometric identities:
$y(x, t) = 2a \sin\left(\frac{(kx – \omega t) + (kx – \omega t + \pi)}{2}\right) \cos\left(\frac{(kx – \omega t) – (kx – \omega t + \pi)}{2}\right)$
$= 2a \sin(kx – \omega t + \frac{\pi}{2}) \cos(-\frac{\pi}{2}) = 0$
Therefore, the correct equation is $y(x, t) = 0$.

Key Concept: Superposition Principle

b) Zero

[Solution Description] According to the principle of superposition, when two waves overlap, their displacements add algebraically. At a rigid boundary, both the incident and reflected waves have equal magnitude but opposite signs due to the phase change of $\pi$. Thus, they cancel each other out, resulting in zero net displacement.

Key Concept: Amplitude Function

d) $A(\phi) = 2a \cos \frac{1}{2}\phi$
[Solution Description] The amplitude function for constructive interference is given by $A(\phi) = 2a \cos \frac{1}{2}\phi$. This formula calculates the combined amplitude based on the phase difference $\phi$.

Key Concept: Wave Propagation

c) The medium oscillates around a fixed point

[Solution Description]

As a wave moves through a medium, the particles of the medium oscillate around their rest position but do not move with the wave itself; this means the medium remains stationary overall while allowing the wave to pass through.

Key Concept: Wave Speed

b) 299,792,458 m/s

[Solution Description] The speed of electromagnetic waves in a vacuum is a standard constant value, specifically $c = 299,792,458 \, \text{ms}^{-1}$.

Key Concept: Concept Integration, Advanced Calculation

c) 15

[Solution Description] The speed of sound in a solid bar can be calculated using $v = \sqrt{\frac{Y}{\rho}}$, where $Y$ is the Young’s modulus and $\rho$ is the density of the material. For the steel rod:
$v_{\text{steel}} = \sqrt{\frac{2.0 \times 10^{11}}{7850}}$
Solving above gives:
$v_{\text{steel}} \approx 5045 \, \text{m/s}$
The given speed of sound in air is 343 m/s. Therefore, the ratio of the speed of sound in steel to that in air is:
$\frac{v_{\text{steel}}}{v_{\text{air}}} = \frac{5045}{343} \approx 14.71$
Hence, the ratio is approximately 15.

Key Concept: Amplitude Variation

c) Amplitude varies from zero at nodes to maximum at antinodes

[Solution Description]

In a standing wave, the amplitude varies from zero at the nodes to maximum at the antinodes. Nodes are points where destructive interference always occurs, resulting in zero displacement, while antinodes are points of maximum constructive interference.

Thus, the correct statement is that the amplitude varies from zero at nodes to maximum at antinodes.

Key Concept: Wave Propagation

b) Longitudinal waves

[Solution Description]

Longitudinal waves can propagate through both solids and fluids because they involve compressions and rarefactions that are possible in all states of matter. Transverse waves, on the other hand, typically only propagate in solids due to their shearing motion.

Key Concept: Real-world Application

b) Longitudinal

[Solution Description] When a metallic bar is struck, longitudinal waves are primarily responsible for sound propagation through the solid material. Transverse waves can also exist but are less significant in terms of speed and energy transfer compared to longitudinal waves in solids. Longitudinal waves require the medium’s elasticity and density for transmission, making them more prominent in metals.

Key Concept: Dependence on Medium

b) Wavelength

[Solution Description]

The speed of transverse waves on a string is determined by properties of the medium, specifically tension and linear mass density. It does not depend on wavelength or frequency; these factors do not alter the speed according to the given formula.

Key Concept: Complex Real-World Scenario, Humidity and Sound Speed

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To evaluate the assertion, we recognize that increased humidity introduces more water vapor into the air, which is less dense than dry air. Thus, the overall density $\rho$ of humid air decreases. According to the formula $v = \sqrt{\frac{P}{\rho}}$, where $v$ is the speed of sound, $P$ is the pressure, and $\rho$ is the density, decreasing $\rho$ increases $v$. Therefore, the Assertion is true.
For the reason, in a closed room, varying humidity does affect air density due to the presence of water vapor. This change in density affects the speed of sound as predicted by the equation $v = \sqrt{\frac{P_{total}}{\rho}}$, where $P_{total}$ accounts for both partial pressures of dry air and water vapor. Hence, the Reason correctly explains the behavior of sound speed in such conditions. Consequently, both the Assertion and the Reason are true, and the Reason is the correct explanation for the Assertion.

Key Concept: Boundary Condition Application

a) $L$
[Solution Description]
The fundamental mode in an open pipe has the wavelength $2L$. The first overtone, also known as the second harmonic, has twice the frequency of the fundamental mode, hence its wavelength is $\frac{L}{1} = L$.

Key Concept: Wave Speed Calculation

b) 5 m/s

[Solution Description] The wave speed $v$ is given by the formula $v = \frac{\omega}{k}$. Substituting the values, we get $v = \frac{10}{2} = 5$ m/s.

Key Concept: Frequency and Period Relationship

a) 0.25 Hz

[Solution Description] Frequency ($\nu$) is the reciprocal of the period (T), so $\nu = \frac{1}{T}$. For $T = 4$ seconds, $\nu = \frac{1}{4} = 0.25$ Hz.

Key Concept: Formula Application

a) 2 m

[Solution Description] To find the position of the first antinode, substitute $n = 0$ into the equation:
$x = (0 + ½) \frac{4}{2} = 1 \times 2 = 2 \text{ m}$
Therefore, the position is 2 meters.

Key Concept: Period Calculation

b) $\frac{\pi}{4}$ seconds
[Solution Description] The period $T$ of a wave is related to its angular frequency $\omega$ by the formula $T = \frac{2\pi}{\omega}$. Substituting $\omega = 8\, \text{rad s}^{-1}$ gives:
$T = \frac{2\pi}{8} = \frac{\pi}{4}\, \text{s}$
Therefore, the period $T$ is $\frac{\pi}{4}$ seconds.