Waves – NEET UG PhysicsPractice Questions & PYQs

When two distinct periodic motions are combined, the resulting oscillation produces a pattern where the beat frequency is equal to the absolute difference between the frequencies of the individual waves. Given periods of 4 s and 3 s, the corresponding frequencies are $f_1 = \frac{1}{4}$ Hz and $f_2 = \frac{1}{3}$ Hz. We determine the beat frequency by calculating $f = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$ Hz. Since the time period is the inverse of this frequency, the resultant period is 12 s.

In wave motion, the relationship between phase difference and path difference is rooted in the fact that one full cycle of the wave, represented by $2\pi$ radians or $360^{\circ}$, corresponds exactly to a distance of one wavelength $\lambda$. A phase difference of $60^{\circ}$ is equivalent to $\frac{\pi}{3}$ radians, allowing the use of the proportional equation $\phi = \frac{2\pi}{\lambda} \Delta x$. Isolating the path difference $\Delta x$ yields $\Delta x = \phi \times \frac{\lambda}{2\pi}$. Substituting the provided phase shift results in $\frac{\pi}{3} \times \frac{\lambda}{2\pi}$, which simplifies to $\frac{\lambda}{6}$.

The intensity of a sound wave is directly proportional to the square of its amplitude and the square of its frequency, expressed by the relationship $I \propto A^{2} f^{2}$. When the amplitude is doubled to $2A$ and the frequency is reduced to $\frac{f}{4}$, the new intensity $I'$ becomes proportional to $(2A)^{2} \times (\frac{f}{4})^{2}$. Calculating this change results in $I' \propto 4A^{2} \times \frac{f^{2}}{16}$, which simplifies to $I' \propto \frac{1}{4} A^{2} f^{2}$. Consequently, the intensity of the sound is reduced to one-fourth of its initial value, representing a decrease by a factor of 4.

The general form of a progressive wave equation is $y = a \sin \left[ 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right) + \phi \right]$, where $a$ denotes amplitude, $T$ denotes time period, and $\lambda$ denotes wavelength. Applying this standard form to the given equation $y = 4 \sin \left[ \pi \left( \frac{t}{5} - \frac{x}{9} \right) + \frac{\pi}{6} \right]$, we rewrite the argument by factoring out $2\pi$ to obtain $y = 4 \sin \left[ 2\pi \left( \frac{t}{10} - \frac{x}{18} \right) + \frac{\pi}{6} \right]$. Through direct comparison between these expressions, the amplitude $a$ is 4 cm, the time period $T$ is 10 s, and the wavelength $\lambda$ is 18 cm.

A harmonic progressive wave is defined by a mathematical relationship that integrates both spatial and temporal variables to describe a disturbance moving through a medium. The equation $y = A \cos(a t - b x + c)$ satisfies this condition, where $a$ acts as the angular velocity, $b$ functions as the wave number, and $c$ represents the initial phase, characterizing a wave that propagates along the positive $x$-direction. In contrast, forms that rely solely on time, such as $y = A \sin \omega t$, describe simple harmonic motion rather than a wave because they lack the necessary spatial dependence to track the displacement of the medium across a distance.

The Doppler effect depends exclusively on the component of the source's velocity that is directed along the line joining the source and the observer, as any motion perpendicular to this axis creates no relative radial velocity. Since the vehicle is moving at an angle of $90^\circ$ relative to the observer, the effective velocity contributing to the frequency shift-given by the term $v_s \cos \theta$-becomes zero because $\cos 90^\circ = 0$. With no radial velocity component, the perceived frequency $n'$ must equal the source frequency $n$. Given that the observer perceives $n' = n + n_1$, substituting $n'$ with $n$ leads directly to the conclusion that $n_1 = 0$.

An open pipe of length $L$ creates standing waves with a fundamental frequency of $f = \frac{v}{2L}$. When submerged in water to half its length, the water surface acts as a reflective barrier, effectively closing one end of the pipe and reducing the air column length to $\frac{L}{2}$. The fundamental frequency of a pipe closed at one end is defined by the formula $f' = \frac{v}{4L'}$, where $L'$ is the length of the air column. Substituting the new effective length $\frac{L}{2}$ into this formula yields $f' = \frac{v}{4(L/2)}$, which simplifies mathematically to $\frac{v}{2L}$. Because this result is identical to the original frequency, the fundamental frequency remains $f$.

When unpolarized light strikes a boundary at Brewster's angle, the reflected light becomes perfectly plane-polarized, whereas the transmitted light remains only partially polarized. The governing relation for this phenomenon is $\mu = \tan\theta_P$, where $\mu$ represents the refractive index of the medium. Given the refractive index of 1.73, the equation becomes $\tan\theta_P = 1.73$, which identifies the polarizing angle as 60 degrees. Consequently, the reflected light is completely polarized, and the angle of reflection is 60 degrees.

The behavior of light intensity through successive polarizers is governed by Malus's Law, which states that transmitted intensity equals the incident intensity multiplied by the squared cosine of the angle between the polarizer's axis and the polarization direction of the light. For this setup, light starts with intensity $I_0$ after the first polarizer. As it passes through a second sheet oriented at $22.5^{\circ}$, the intensity becomes $I_1 = I_0 \cos^2(22.5^{\circ})$.

Since the third polarizer is crossed with the first, its axis is oriented at $90^{\circ} - 22.5^{\circ} = 67.5^{\circ}$ relative to the second polarizer. The final intensity is therefore $I_2 = I_1 \cos^2(67.5^{\circ})$, which simplifies to $I_0 \cos^2(22.5^{\circ}) \sin^2(22.5^{\circ})$ because the cosine of $67.5^{\circ}$ is equal to the sine of $22.5^{\circ}$. Using the double-angle identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, the expression becomes $I_0 \left( \frac{\sin(45^{\circ})}{2} \right)^2$, resulting in a final value of $\frac{I_0}{8}$.

In Young's double slit experiment, the spacing between consecutive interference fringes is defined by the formula $\beta = \frac{\lambda D}{d}$, where $\lambda$ represents the wavelength of light, $D$ is the distance from the slits to the screen, and $d$ is the separation between the slits. To determine the fringe width, we first standardize all units to meters, noting that the wavelength is $600 \times 10^{-9}$ meters, the screen distance is $1$ meter, and the slit separation of $1.5$ millimeters is equivalent to $1.5 \times 10^{-3}$ meters. Substituting these values into the expression provides
$\beta = \frac{600 \times 10^{-9} \times 1}{1.5 \times 10^{-3}}$
Performing the calculation, dividing $600$ by $1.5$ yields $400$, while the powers of ten simplify to $10^{-6}$, resulting in a final fringe separation of $4 \times 10^{-4}$ meters.

For a stable interference pattern to manifest, the superimposing waves must originate from coherent sources, which necessitates that they operate at an identical frequency. Waves represented by $y = a \sin \omega t$ and $y = a \sin (\omega t - \phi)$ fulfill this requirement because they both oscillate at the same angular frequency, $\omega$. Conversely, the expressions involving $2\omega$ and $3\omega$ represent different frequencies, which precludes the establishment of the consistent phase relationship required for observable interference.

Any traveling wave is characterized by the general equation $y = A \sin (\omega t - kx)$, where $\omega$ stands for the angular frequency and $k$ denotes the wave number. Distributing the factor $\frac{2\pi}{\lambda}$ across the expression inside the sine function, we rewrite the given equation as $y = C \sin (\frac{2\pi a}{\lambda} t - \frac{2\pi}{\lambda} x)$, which allows us to directly identify the coefficient of time as the angular frequency. Setting $\omega = \frac{2\pi a}{\lambda}$ and utilizing the fundamental relationship $\omega = 2\pi f$, we equate $2\pi f = \frac{2\pi a}{\lambda}$, which simplifies to $f = \frac{a}{\lambda}$.

For a closed organ pipe, the frequency of the $n^{\text{th}}$ overtone corresponds to the $(2n+1)^{\text{th}}$ harmonic, given by the expression $\frac{(2n+1)v}{4L_c}$, which means the $4^{\text{th}}$ overtone resonates at $\frac{9v}{4L_c}$. An open organ pipe, which supports all harmonics, has a frequency for the $n^{\text{th}}$ overtone defined by $(n+1)\frac{v}{2L_o}$, making the $3^{\text{rd}}$ overtone equal to $\frac{4v}{2L_o}$. Equating the frequencies of both pipes, we have $\frac{9v}{4L_c} = \frac{4v}{2L_o}$. Canceling the velocity of sound $v$ from both sides and simplifying the fractions results in $\frac{9}{4L_c} = \frac{2}{L_o}$, which determines the ratio of the length of the closed pipe to the length of the open pipe to be $\frac{L_c}{L_o} = \frac{9}{8}$.

The fundamental frequency of any pipe is dictated by the speed of sound and the length of the air column, which determines the specific wavelength of the standing wave produced. An open pipe operates as an open-open system where the fundamental mode corresponds to exactly half a wavelength spanning its length $L$, resulting in a frequency defined as $f_{open} = \frac{v}{2L}$. Conversely, a closed pipe functions as an open-closed system, where boundary conditions require a node at the closed end and an antinode at the open end, allowing only a quarter of a wavelength to exist within the same length $L$, which produces a fundamental frequency of $f_{closed} = \frac{v}{4L}$. Comparing these two values reveals that the ratio of the open pipe frequency to the closed pipe frequency is $\frac{v/2L}{v/4L}$, which simplifies to $2:1$.

The fundamental frequency of an organ pipe is determined by the speed of sound within the gas, and because the speed of sound is directly proportional to the square root of the absolute temperature, the frequency likewise follows the relationship $f \propto \sqrt{T}$. Converting the given temperatures to Kelvin by adding 273 yields an initial temperature of $300 \text{ K}$ and a final temperature of $363 \text{ K}$. Using the proportionality ratio
$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$
substituting the known values gives
$\frac{f_2}{400} = \sqrt{\frac{363}{300}}$
which simplifies to
$\frac{f_2}{400} = \sqrt{1.21} = 1.1$
Multiplying 400 by 1.1 results in a final resonant frequency of 440 Hz.

The speed of a transverse wave propagating along a stretched string is determined by the tension $T$ and the linear mass density $\mu$ according to the relation $v = \sqrt{\frac{T}{\mu}}$. Because the linear mass density remains constant, the wave speed is directly proportional to the square root of the tension. When the initial tension $T$ is doubled to a final value of $2T$, the ratio of the initial speed $v_i$ to the final speed $v_f$ is calculated as follows:

$\frac{v_i}{v_f} = \frac{\sqrt{\frac{T}{\mu}}}{\sqrt{\frac{2T}{\mu}}} = \frac{1}{\sqrt{2}}$

This mathematical relationship shows that the ratio of the initial speed to the final speed is $1:\sqrt{2}$.

Beats occur at a frequency defined by the absolute difference between two notes, calculated as $|f_A - f_B|$, while a string's frequency itself drops when its tension is decreased. Since string A has a frequency of 530 Hz and creates 6 Hz beats, the starting frequency of string B must be either 524 Hz or 536 Hz. Because reducing the tension of string B causes its frequency to drop, we evaluate how this change shifts the beat frequency relative to string A. If the starting frequency were 536 Hz, a decrease in tension would bring the note closer to 530 Hz, thereby reducing the number of beats; however, since the beat frequency rises to 7 Hz, the note must be shifting further away from 530 Hz, which confirms the original frequency was 524 Hz.

In a resonance column open at one end and closed at the other, standing waves form such that the difference between any two consecutive resonant lengths always equals half the wavelength of the sound wave, expressed as $\Delta l = \frac{\lambda}{2}$. Because the frequency $f$ relates to the wavelength and wave speed $v$ by the equation $v = f\lambda$, we can determine the speed of sound by calculating the wavelength from the experimental measurements.

Using the resonance lengths of 9.75 cm and 31.25 cm, the difference is 21.5 cm, or 0.215 m. Since this difference represents half a wavelength, we find the full wavelength using the equation $\lambda = 2 \times 0.215 = 0.43$ m. Substituting this wavelength and the frequency of 800 Hz into the wave speed formula, the speed is $v = 800 \times 0.43 = 344$ m/s.

In a closed resonance tube, the distance between two successive resonance positions corresponds to exactly half of the sound wavelength. This fundamental standing wave behavior implies that the difference between the two measured lengths, 73 cm and 20 cm, is equal to $\lambda/2$, which allows us to determine the wavelength of the sound produced by the tuning fork.

Because the velocity of a wave is the product of its frequency and wavelength, represented by $v = f\lambda$, we can express the relationship for this setup as $v = 2f(L_2 - L_1)$ where the resonance lengths are converted into meters. By substituting the frequency of 320 Hz and the length difference of 0.53 meters into the equation $v = 2 \times 320 \times 0.53$ we calculate the velocity to be 339.2 m/s, which rounds to 339 m/s.

Sound waves trapped in organ pipes create specific resonances determined by the pipe's geometry, where the fundamental frequency for an open pipe is defined as $f = \frac{v}{2L_o}$, and the third harmonic for a closed pipe is generated at $f = \frac{3v}{4L_c}$.

Equating these two frequencies allows us to solve for the unknown length of the open pipe, $L_o$. Setting the expressions equal, $\frac{v}{2L_o} = \frac{3v}{4L_c}$, we can cancel the wave speed $v$ from both sides to obtain $\frac{1}{2L_o} = \frac{3}{4L_c}$, which rearranges to $L_o = \frac{4L_c}{6}$, or $L_o = \frac{2L_c}{3}$. Substituting the given closed pipe length of $20$ cm yields $L_o = \frac{2 \times 20}{3}$, which results in approximately $13.33$ cm.