Waves – NEET UG PhysicsPractice Questions & PYQs

The fundamental frequency of a vibrating string is inversely proportional to its length when the tension and linear mass density remain constant, expressed by the relationship $\upsilon = \frac{k}{l}$ where $k$ is a constant. Because the total length of the original string is simply the sum of its individual parts, we have the geometric condition $l = l_1 + l_2 + l_3$. By substituting the equivalent expressions $l = \frac{k}{\upsilon}$, $l_1 = \frac{k}{\upsilon_1}$, $l_2 = \frac{k}{\upsilon_2}$, and $l_3 = \frac{k}{\upsilon_3}$ into this sum, we obtain $\frac{k}{\upsilon} = \frac{k}{\upsilon_1} + \frac{k}{\upsilon_2} + \frac{k}{\upsilon_3}$. Canceling the common constant $k$ from both sides directly yields the result that the reciprocal of the original frequency is equal to the sum of the reciprocals of the segment frequencies.

The wave displacement for a simple harmonic wave is described by the standard equation $y=A \sin(\omega t - k x)$, where $A$ is the amplitude, $\omega$ is the angular frequency, and $k$ is the wave number. For the given equation $y=3\sin(\frac{\pi}{2})(50t-x)$, we distribute the constant $\frac{\pi}{2}$ to rewrite the function as $y=3 \sin(25\pi t - \frac{\pi}{2} x)$. Comparing this to the standard form, we identify the amplitude as $A=3$, the angular frequency as $\omega=25\pi$, and the wave number as $k=\frac{\pi}{2}$.

The wave velocity $v$ is calculated using the relationship $v = \frac{\omega}{k}$, which yields $v = \frac{25\pi}{\pi/2} = 50 \text{ m/s}$. To find the maximum particle velocity, we derive the displacement with respect to time to get $v_p = \frac{\partial y}{\partial t} = A\omega\cos(\omega t - kx)$. The maximum particle velocity $(v_p)_{\text{max}}$ is therefore $A\omega$, which equals $3 \times 25\pi = 75\pi \text{ m/s}$. The ratio of the maximum particle velocity to the wave velocity is $\frac{75\pi}{50}$, which simplifies to $\frac{3}{2}\pi$.

To compare the phases of two waves, we must express both using the same trigonometric function, such as converting the cosine term into a sine term using the identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$. Given the second equation $y_2 = a \cos(\omega t + kx)$, we rewrite it as $y_2 = a \sin(\omega t + kx + \frac{\pi}{2})$, where $\frac{\pi}{2}$ is approximately 1.57 radians. Subtracting the phase of the first wave, $\omega t + kx + 0.57$, from the phase of the second wave, $\omega t + kx + 1.57$, yields a phase difference of $1.57 - 0.57$, which results in 1.0 radian.

The fundamental frequency of a vibrating string is described by $v = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, which indicates that frequency is proportional to the square root of tension. Analyzing the relationship between small changes in these variables results in $\frac{\Delta v}{v} = \frac{1}{2} \frac{\Delta T}{T}$.

Rearranging this expression to solve for the fractional change in tension gives $\frac{\Delta T}{T} = 2 \frac{\Delta v}{v}$. Substituting the initial frequency of $600$ Hz and the beat frequency of $6$ Hz, which represents the change in frequency, the calculation becomes $\frac{2 \times 6}{600}$. This yields a final value of $0.02$.

The frequency of a sound wave is a constant property defined by the source and does not change when the wave transitions between different media, which establishes a direct proportionality between wave speed and wavelength according to the relation $v = f \lambda$. Since the sound moves from air with a speed of 350 m/s into brass where the speed is 3500 m/s, the wave speed increases by a factor of 10. Because the frequency remains fixed throughout this transition, the wavelength must increase by this same factor to keep the equation balanced.

The wave velocity describes the speed of the pattern moving through the medium, calculated as $v = \frac{\omega}{k}$, while the maximum particle velocity represents the peak speed of any single particle oscillating in place, given by $v_{max} = A\omega$. By equating these two speeds, we get the relationship $\frac{\omega}{k} = A\omega$, which simplifies to $\frac{1}{k} = A$. Substituting the standard identity for wave number, $k = \frac{2\pi}{\lambda}$, transforms the expression into $\frac{\lambda}{2\pi} = A$, which reveals that the wavelength must be $\lambda = 2\pi A$.

The frequency of a piano string can be determined using the phenomenon of beats, where the beat frequency is defined by the relation $f_{\text{beat}} = |f_{\text{string}} - f_{\text{fork}}|$. Given the tuning fork frequency of 512 Hz and a beat frequency of 4 Hz, the initial frequency of the piano string must be either 516 Hz or 508 Hz. Increasing the tension in a vibrating string always results in an increase in its frequency.

If the string initially vibrated at 516 Hz, adding tension would cause the frequency to rise further above 512 Hz, thereby increasing the beat frequency. Since the problem explicitly states that the beat rate decreased to 2 Hz, the string must have started at 508 Hz. As tension is applied, the frequency moves from 508 Hz toward 512 Hz, effectively reducing the difference between the two sources and lowering the beat frequency to the observed value.

The Doppler effect explains how the frequency of a sound wave shifts due to relative motion between the source and the observer. In this situation, the sound travels from the car to the hill and reflects back, meaning the Doppler effect occurs twice: once when the sound reaches the hill, and again when the reflected sound reaches the driver.

Consider the sound travelling from the car towards the hill. The car acts as the source moving at 30 m/s towards a stationary observer, the hill. The frequency $f_1$ reaching the hill is calculated using the formula $f_1 = f \left( \frac{v}{v - v_s} \right)$, where $f$ is 600 Hz and $v$ is 330 m/s. Substituting these values into the equation, $f_1 = 600 \left( \frac{330}{330 - 30} \right) = 600 \left( \frac{330}{300} \right)$, which equals 660 Hz.

Next, the hill acts as a stationary source emitting this sound at 660 Hz, and the driver acts as an observer moving at 30 m/s towards the hill. The final frequency $f_2$ heard by the driver is given by the formula $f_2 = f_1 \left( \frac{v + v_d}{v} \right)$. Plugging in the values, $f_2 = 660 \left( \frac{330 + 30}{330} \right) = 660 \left( \frac{360}{330} \right)$. Simplifying this expression yields 720 Hz.

The general form for a wave traveling in the positive $x$-direction is expressed as $y = A \sin(kx - \omega t)$, where $A$ denotes amplitude, $k$ is the wave number, and $\omega$ is the angular frequency. With an amplitude of 2 cm converted to $0.02 \text{ m}$, we first identify the wavelength using the provided spatial information, where five complete waves in 4 m leads to a wavelength of $\lambda = 0.8 \text{ m}$. The wave number is calculated as $k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.8} = 7.85 \text{ m}^{-1}$. Simultaneously, the temporal frequency is derived from the wave speed of $128 \text{ m/s}$ using $f = \frac{v}{\lambda} = \frac{128}{0.8} = 160 \text{ Hz}$, which results in an angular frequency of $\omega = 2\pi f = 2\pi \times 160 \approx 1005 \text{ rad/s}$. Substituting these parameters into the standard wave equation yields $y = 0.02 \sin(7.85x - 1005t)$.

The fundamental frequency of a vibrating string is defined by the relationship $f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$, where $l$ is the length of the string, $T$ is the tension, and $\mu$ is the linear mass density. Because both strings are subjected to identical tension and possess the same mass per unit length, the difference in their frequencies is determined entirely by the difference in their lengths.

Given the tension of $20$ N and a linear mass density of $0.001$ kg/m, the common velocity term $\sqrt{\frac{T}{\mu}}$ evaluates to $\sqrt{20000}$, which is approximately $141.42$ m/s. Substituting the specific lengths of $0.516$ m and $0.491$ m into the frequency formula, the first string vibrates at $f_1 = \frac{141.42}{2 \times 0.516} \approx 137.0$ Hz, while the second string vibrates at $f_2 = \frac{141.42}{2 \times 0.491} \approx 144.0$ Hz. The number of beats per second produced when both strings are played simultaneously is the absolute difference between these two frequencies, $144.0 - 137.0$, resulting in $7$ beats.

A traveling wave is mathematically represented by the function $y = A \sin(kx - \omega t)$, where the sign between the spatial term $kx$ and the temporal term $\omega t$ dictates the direction of motion; specifically, a negative sign indicates propagation in the positive x-direction. Comparing the given equation to this standard form, we identify the angular wavenumber $k = 10\pi$ and angular frequency $\omega = 2\pi$. We determine the wavelength $\lambda$ using the relationship
$k = \frac{2\pi}{\lambda}$
which equates to $10\pi = \frac{2\pi}{\lambda}$, resulting in $\lambda = 0.2$ m. Similarly, the frequency $f$ is calculated using the equation
$\omega = 2\pi f$
which leads to $2\pi = 2\pi f$, establishing that the frequency is $f = 1$ Hz.

The principle of superposition dictates that the resultant amplitude of interfering waves is the algebraic sum or difference of their individual amplitudes, which are related to the square root of their respective intensities. For constructive interference, the maximum intensity is $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 = I_1 + I_2 + 2\sqrt{I_1I_2}$, while for destructive interference, the minimum intensity is $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 = I_1 + I_2 - 2\sqrt{I_1I_2}$. By summing these two expressions, the interference cross-terms cancel perfectly. This addition results in a total value of $2(I_1 + I_2)$.

The velocity of a particle undergoing simple harmonic motion is obtained by differentiating its displacement equation $x = a \sin(\omega t + \pi/6)$ with respect to time, which yields $v = a \omega \cos(\omega t + \pi/6)$. Given that the maximum velocity is $a \omega$, we find the required time by setting the expression for velocity equal to $a \omega / 2$, leading to $\cos(\omega t + \pi/6) = 1/2$. Because the cosine function equals $1/2$ at $\pi/3$ radians, the phase must satisfy $\omega t + \pi/6 = \pi/3$, which simplifies to $\omega t = \pi/6$. Substituting the angular frequency $\omega = 2\pi/T$ into this equation gives $(2\pi/T)t = \pi/6$, confirming the elapsed time is $T/12$.

Beats occur when two sound waves of slightly different frequencies interfere, creating a periodic variation in intensity where the beat frequency is the absolute difference between them. This frequency is determined using the relationship $f = \frac{v}{\lambda}$. Applying this to the given values, dividing the velocity 330 m/s by the wavelengths 5.0 m and 5.5 m yields individual frequencies of 66 Hz and 60 Hz, respectively. The resulting beat frequency is calculated as $66 - 60 = 6$ beats per second.

Light waves function as electromagnetic waves with a transverse nature, which allows them to propagate efficiently through a vacuum without needing any material medium. Sound waves, however, propagate as longitudinal waves that must interact with particles in a medium to move, meaning they lack the ability to travel through a vacuum.

The phenomenon of beats arises when two waves with slightly differing frequencies overlap, creating a periodic variation in intensity defined by the absolute difference between their respective frequencies, expressed as $f_{beat} = |f_{2} - f_{1}|$. Since the general equation for a wave is $y = A \sin(\omega t)$ where the angular frequency $\omega$ equals $2\pi f$, we can extract the individual frequencies for each tuning fork. For the first wave, setting $500\pi = 2\pi f_{1}$ results in $f_{1} = 250$ Hz, while setting $506\pi = 2\pi f_{2}$ for the second wave yields $f_{2} = 253$ Hz. The interference between these two forks generates a beat frequency of $|253 - 250| = 3$ beats per second. Because the request asks for the number of beats per minute, we multiply this rate by 60, resulting in a total of 180 beats per minute.

The velocity of a traveling wave is defined by the ratio of its angular frequency to its wave number, which are the coefficients associated with time and position in the standard equation $y(x, t) = A \sin(kx - \omega t + \phi)$. By comparing this standard structure to the expression $y(x, t) = 8.0 \sin(0.5 \pi x - 4 \pi t - \pi/4)$, we determine that the wave number is $k = 0.5 \pi$ and the angular frequency is $\omega = 4 \pi$. The speed $v$ is computed as:

$v = \frac{\omega}{k} = \frac{4 \pi}{0.5 \pi} = 8 \text{ m/s}$

The wave moves at a speed of $8 \text{ m/s}$.

The reverberation time of an enclosure is governed by Sabine's formula, which establishes the proportionality $T \propto \frac{V}{S}$, where $V$ represents the volume and $S$ represents the total surface area of the room. When every linear dimension of a room is doubled, the volume increases by a factor of $2^3 = 8$, while the total surface area increases by a factor of $2^2 = 4$. Substituting these scaling factors into the proportionality relation reveals that the new reverberation time $T_{2}$ relates to the original time $T_{1}$ by a factor of $\frac{8}{4} = 2$. Given that the original reverberation time is 1 second, the new duration is $2 \times 1 = 2$ seconds.

Sound energy radiating from a point source into a non-absorbing medium spreads over the surface of expanding spheres, which leads to the inverse square law. Because the area of these wavefronts increases with the square of the distance, the intensity $I$ is inversely proportional to the square of the distance $r$ from the source, as shown by the relationship $I \propto \frac{1}{r^2}$.

Since point $P$ is at a distance of 2 m and point $Q$ is at a distance of 3 m, the ratio of the intensity at $P$ to the intensity at $Q$ is found by comparing their inverse squares. This calculation yields the following result:

$\frac{I_P}{I_Q} = \frac{3^2}{2^2} = \frac{9}{4}$

This scenario involves a two-stage Doppler shift because the sound wave encounters the moving vehicle twice. First, the car acts as a source emitting frequency $f$ toward a stationary cliff, which receives a shifted frequency $f' = f \left( \frac{v}{v - v_{car}} \right)$, where $v_{car}$ is the speed of the car. Since the cliff reflects this sound, it acts as a stationary source emitting frequency $f'$ back toward the car, which is now the observer moving toward that source. The driver consequently hears a final frequency $f''$ calculated by $f'' = f' \left( \frac{v + v_{car}}{v} \right)$.

Substituting the first expression for $f'$ into the equation for $f''$ allows us to equate the observed reflected frequency to $2f$, resulting in $2f = f \left( \frac{v}{v - v_{car}} \right) \left( \frac{v + v_{car}}{v} \right)$. Canceling the common terms simplifies the relationship to $2 = \frac{v + v_{car}}{v - v_{car}}$. Expanding this equation gives $2(v - v_{car}) = v + v_{car}$, which further simplifies to $2v - 2v_{car} = v + v_{car}$. Solving for $v_{car}$ yields $3v_{car} = v$, indicating that the speed of the car is $\frac{v}{3}$.