Waves – NEET UG PhysicsPractice Questions & PYQs

When a string is fixed at one end, a wave reflecting off that boundary interferes with the incoming wave to form a stationary wave pattern where the fixed end consistently acts as a node. In this standing wave, the physical distance between any two consecutive nodes is equal to $\lambda / 2$. Since the distance from the fixed end node to the next node is 10 cm, the wavelength is 20 cm, which converts to 0.2 m. Using the fundamental wave equation $v = f \lambda$, we multiply the frequency of 100 Hz by the wavelength of 0.2 m to find that the speed of the wave is 20 m/s.

In a sinusoidal wave, the motion from maximum displacement to the equilibrium position corresponds to exactly one-quarter of a complete oscillation, or $\frac{T}{4}$. With this time interval given as $0.170\text{ s}$, we find the total period by multiplying this value by four to obtain $T = 0.68\text{ s}$. Since the frequency is the reciprocal of the time period, calculating $f = \frac{1}{T}$ yields $f = \frac{1}{0.68\text{ s}}$, which results in $1.47\text{ Hz}$.

Beat frequency arises from the superposition of two sound waves of slightly different frequencies, where the number of beats per second is equal to the magnitude of the difference between the two frequencies. When a source with an unknown frequency $\nu$ interacts with a known frequency $F$ at a rate of $n$ beats per second, the unknown frequency is expressed by the relation $\nu = F \pm n$.

The first scenario involves a frequency $\nu$ generating 5 beats per second with a $200\,\text{Hz}$ source, which implies that $\nu$ must be either $195\,\text{Hz}$ or $205\,\text{Hz}$. In the second scenario, the second harmonic $2\nu$ creates 10 beats per second when paired with a $420\,\text{Hz}$ source, meaning $2\nu$ is either $410\,\text{Hz}$ or $430\,\text{Hz}$. Dividing these values by two indicates that $\nu$ could be either $205\,\text{Hz}$ or $215\,\text{Hz}$. Evaluating the possibilities from both situations shows that $205\,\text{Hz}$ is the only frequency that satisfies both conditions simultaneously.

Beats occur when two sound waves with slightly different frequencies interfere, resulting in a beat frequency equal to the difference between the two source frequencies. Since the frequency of a wave is determined by the ratio of velocity to wavelength, expressed as $f = \frac{v}{\lambda}$, the beat frequency of 12 Hz is the difference between the frequencies associated with wavelengths of 50 cm and 51 cm. We define this relationship as:

$v \left( \frac{1}{50} - \frac{1}{51} \right) = 12$

Simplifying the term inside the parentheses leads to a fraction of $\frac{1}{2550}$, so the equation becomes $v \times \frac{1}{2550} = 12$. Solving for the velocity yields 30600 cm/s, which converts to 306 m/s in standard units.

The fundamental frequency of a sonometer wire is inversely proportional to its length when tension and linear mass density remain constant, a relationship expressed as $f \propto \frac{1}{l}$. Because the required fundamental frequencies for the three segments are in the ratio $1 : 2 : 3$, the lengths of these segments must follow the inverse ratio of $\frac{1}{1} : \frac{1}{2} : \frac{1}{3}$. Multiplying these terms by their least common multiple simplifies this proportion to $6 : 3 : 2$.

With a total wire length of $110$ cm, the sum of these ratio components is $6 + 3 + 2 = 11$. Dividing the total length by this sum gives $110 / 11 = 10$ cm per unit of the ratio. Consequently, the lengths of the three segments are $6 \times 10 = 60$ cm, $3 \times 10 = 30$ cm, and $2 \times 10 = 20$ cm. To achieve this division, the first bridge must be positioned $60$ cm from the starting end, and the second bridge must be placed $30$ cm beyond that, which is at a distance of $90$ cm from the first end.

The fundamental wave equation $\lambda = v / \nu$ describes the relationship between the speed of sound, its frequency, and its wavelength. Converting the provided scanner parameters into standard SI units yields a frequency of $\nu = 4.2 \times 10^6$ Hz and a sound speed of $v = 1.7 \times 10^3$ m/s. Substituting these values into the wave equation gives $\lambda = (1.7 \times 10^3) / (4.2 \times 10^6)$. Completing this calculation results in $\lambda \approx 4 \times 10^{-4}$ m.

The fundamental frequency of an air column is dictated by the pipe's length and its boundary conditions, which are open at both ends in the initial state $f = \frac{v}{2l}$. Placing half the tube in water forces a change in boundary conditions, transforming it into a closed pipe with an effective length of $\frac{l}{2}$. Calculating the new frequency for this closed pipe gives $f' = \frac{v}{4(\frac{l}{2})}$, which simplifies back to $\frac{v}{2l}$, proving that the fundamental frequency remains $f$.

The frequency of a vibrating stretched string forming standing waves is determined by its physical dimensions and the mode of vibration, which is described by the relation $f = \frac{n v}{2 L}$. In this formula, n represents the number of segments, v is the wave velocity, and L is the length of the string. Substituting the given values of 5 segments, 20 m/s velocity, and 10 m length into the formula yields $f = \frac{5 \times 20}{2 \times 10}$, which results in a final frequency of 5 Hz.

A stationary wave is created by the superposition of two progressive waves traveling in opposite directions, and its mathematical structure is generally given by $y = A \sin(\omega t) \cos(kx)$, where $\omega$ represents the angular frequency and $k$ denotes the wave number. By comparing this standard form to the equation $y = A \sin(100 t) \cos(0.01 x)$, we identify the angular frequency as 100 and the wave number as 0.01.

The velocity of the wave is determined by the relationship $v = \frac{\omega}{k}$ which relates the temporal oscillation to the spatial periodicity. Substituting the extracted values into this formula, we calculate the speed as $v = \frac{100}{0.01}$ resulting \in a value of 10000 m/s, which can also be represented as $10^{4}$ m/s.

The relationship between path difference $\Delta x$ and phase difference $\Delta \phi$ is expressed by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$, where $\lambda$ represents the wavelength. Rearranging this expression to solve for wavelength gives the equation $\lambda = \frac{2 \pi \Delta x}{\Delta \phi}$. Substituting the given path difference of 0.4 m and phase difference of $1.6 \pi$ leads to $\lambda = \frac{2 \pi \times 0.4}{1.6 \pi}$, which simplifies to 0.5 m. Using the fundamental wave equation $v = f \lambda$ to find the frequency $f$, dividing the velocity of 330 m/s by the calculated wavelength of 0.5 m results in 660 Hz.

Mechanical waves propagate through a medium by transporting energy and linear momentum without causing the net displacement of the material itself. As particles of the medium oscillate about their equilibrium positions, they possess velocity and mass, which enables the transfer of both energy and momentum, characterized by $p = mv$, through the system. While the disturbance travels forward, the constituent particles remain confined to their local region, ensuring that only the wave energy and momentum are carried along the direction of propagation rather than the physical matter.

In a pipe closed at one end, the third overtone corresponds to the fourth harmonic, characterized by a frequency of $4n$. This vibrational state requires the air column to establish a standing wave pattern that satisfies the boundary conditions at both the open and closed ends. Consequently, this specific mode manifests as a wave structure containing exactly four nodes and four antinodes.

In a traveling wave, the velocity of an individual particle oscillating about its mean position is found by taking the time derivative of its displacement equation. Differentiating the function $y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ with respect to time gives the particle velocity, and since the maximum value of the cosine component is unity, the peak particle velocity is $v_{max} = y_0 \frac{2\pi v}{\lambda}$. Equating this maximum velocity to twice the wave velocity $v$ as specified by the problem, we arrive at the following relationship:

$y_{0} \frac{2 \pi v}{\lambda} = 2v$

Dividing both sides by the wave velocity $v$ and rearranging the expression to solve for the wavelength yields the final result of $\lambda = \pi y_{0}$.

In a standing wave pattern, the distance between any adjacent node and antinode is defined as $\frac{\lambda}{4}$. Given the configuration consists of three nodes and two antinodes, the distance between the two fixed atomic boundaries spans two full wave loops. This total path length is equivalent to four node-to-antinode segments, resulting in the equation $4 \times \frac{\lambda}{4} = \lambda$. Since the atoms are separated by 1.21 Å, the wavelength of the standing wave is equal to this value, which is 1.21 Å.

Beats occur when two waves with slightly different frequencies interfere, resulting in a characteristic periodic fluctuation in loudness. While the amplitudes of the waves determine the specific depth of these intensity variations and their phase relationship evolves naturally due to the frequency mismatch, the existence of different frequencies is the solitary fundamental requirement for this phenomenon to occur.

The phase difference between any two points in a travelling harmonic wave is determined by the product of the wave number and the separation distance, represented by the relationship $\Delta \phi = k \Delta x$. In the wave equation $y(x, t)=2.0 \cos 2 \pi(10 t-0.0080 x+0.35)$, the coefficient associated with the spatial variable $x$ identifies the wave number as $k = 2\pi \times 0.0080$ rad/cm. Since the wave equation parameters are expressed in centimeters, the provided separation distance of $0.5$ m must be converted to $50$ cm to maintain dimensional consistency. Multiplying the wave number by this distance results in $\Delta \phi = (2\pi \times 0.0080) \times 50$, which simplifies to a phase difference of $0.8 \pi$ rad.

Any progressive wave is defined by the standard equation $y = A \sin(kx + \omega t)$, where $k$ represents the wave number given by the relationship $k = \frac{2\pi}{\lambda}$. By observing the provided equation $y = 0.0015 \sin(62.4x + 316t)$, we can identify the wave number $k$ as $62.4$. Setting $\frac{2\pi}{\lambda}$ equal to $62.4$ allows us to isolate the wavelength, resulting in the calculation $\lambda = \frac{2\pi}{62.4}$, which simplifies to $0.1$ unit.

Beats arise from the superposition of two sound waves, where the beat frequency is the absolute difference between their individual frequencies, and the second harmonic is defined as twice the fundamental frequency. When the unknown frequency $f$ generates 5 beats per second with a source of 100 sec^-1, the possible values for $f$ are 105 sec^-1 or 95 sec^-1. The second harmonic, $2f$, would therefore be either 210 sec^-1 or 190 sec^-1. Given that this harmonic produces 5 beats per second when sounded with a 205 sec^-1 source, we determine which value satisfies the condition $|2f - 205| = 5$. Evaluating these possibilities, $|210 - 205| = 5$ while $|190 - 205| = 15$, which confirms that the frequency of the source is 105 sec^-1.

In an open-ended pipe, the number of nodes in a standing wave is equal to the harmonic number, which means the 5th harmonic has $n = 5$ nodes. For a pipe closed at one end, standing waves are restricted to odd harmonics, where the count of nodes follows a specific pattern: the 1st harmonic has 1 node, the 3rd has 2, the 5th has 3, the 7th has 4, and the 9th harmonic has $m = 5$ nodes. The ratio of these two values is therefore $\frac{n}{m} = \frac{5}{5} = 1$.

The Doppler effect describes how the perceived wavelength of radiation changes when a source is in motion relative to an observer. For an approaching source, the radiation undergoes a blue shift, and the magnitude of this change in wavelength is proportional to the ratio of the source's velocity to the speed of light. This relationship is defined by the formula $\Delta \lambda = \lambda \frac{v}{c}$, where $\Delta \lambda$ is the change in wavelength, $\lambda$ is the initial wavelength, $v$ is the velocity of the source, and $c$ is the speed of light.

To calculate the specific change for this star, we substitute the known values of $\lambda = 5000\,\text{\AA}$, $v = 1.5 \times 10^{6}\,\text{m/s}$, and the universal constant $c = 3 \times 10^{8}\,\text{m/s}$ into the equation. The calculation becomes:

$\Delta \lambda = 5000 \times \frac{1.5 \times 10^{6}}{3 \times 10^{8}}$

Simplifying this expression results in a value of 25 Å, representing the total shift in the radiation wavelength received on Earth.