Waves – NEET UG PhysicsPractice Questions & PYQs

The shift in sound frequency resulting from the relative motion between a source and an observer is described by the Doppler effect. When cars move toward each other, the sound waves are compressed, causing the listener to perceive a higher pitch than what the source originally emits. This phenomenon is governed by the formula $f_{app} = f_0 \left( \frac{v + v_{observer}}{v - v_{source}} \right)$, where the source frequency $f_0$ is 400 Hz, the speed of sound $v$ is 340 m/s, the source velocity $v_{source}$ is 22 m/s, and the observer velocity $v_{observer}$ is 16.5 m/s. Entering these values yields $f_{app} = 400 \left( \frac{340 + 16.5}{340 - 22} \right)$, which simplifies to $f_{app} = 400 \left( \frac{356.5}{318} \right)$ for a final value of 448 Hz.

In a tube closed at one end, standing waves are formed such that the allowed harmonic frequencies are restricted to odd integer multiples of the fundamental frequency. Because consecutive harmonics in this system are separated by an interval equal to twice the fundamental frequency, the difference between any two adjacent resonant frequencies provides a direct way to calculate that base value.

Given that 220 Hz and 260 Hz are the nearest consecutive harmonics, their difference of 40 Hz represents the value of $2f_0$, where $f_0$ is the fundamental frequency. By solving the equation $2f_0 = 40$ Hz, we determine that the fundamental frequency of the system is 20 Hz.

Beats arise when waves of slightly differing frequencies overlap, leading to periodic fluctuations in sound intensity measured by the absolute difference between those frequencies. In a system with three waves characterized by frequencies of $n-1$, $n$, and $n+1$, the resulting beat frequency is determined by the total range covered by these interfering waves. Calculating the difference between the maximum frequency, $n+1$, and the minimum frequency, $n-1$, yields the expression $(n+1) - (n-1)$, which simplifies to 2. Consequently, the superposition of these waves results in 2 beats per second.

The reflection of sound off a cliff is effectively a process where the cliff acts as a stationary boundary that echoes back the sound it receives. Because the siren moves towards the cliff at 15 m/s, the wavefronts are compressed in the direction of the reflection, causing the cliff to receive a frequency higher than the original 800 Hz. Since the observer remains stationary while the reflected sound waves travel back towards them, the frequency they hear corresponds to this compressed value defined by the Doppler effect. The relationship between the original frequency $f$, the velocity of sound $v = 330$ m/s, and the source velocity $u_S = 15$ m/s is represented by the formula:

$f^{\prime} = \left( \frac{v}{v - u_S} \right) f$

Substituting the known values into this equation allows for the determination of the final frequency heard by the observer:

$f^{\prime} = \left( \frac{330}{330 - 15} \right) 800 = \left( \frac{330}{315} \right) 800 \approx 838 \, \text{Hz}$

This results in an observed echo frequency of 838 Hz.

Resonant frequencies in organ pipes are dictated by the physical length of the pipe and the boundary conditions at each end. For an open organ pipe of length $l_0$, the standing waves are governed by the equation $f_n = \frac{n V}{2 l_0}$, where $n$ represents the harmonic number. Since the second overtone corresponds to the third harmonic, the frequency is $f_{open} = \frac{3 V}{2 l_0}$.

In a closed pipe of length $L$, the sound wave is constrained to odd harmonics, defined by $f_n = \frac{n V}{4 L}$. The first overtone for this pipe occurs at the third harmonic, where $n = 3$, giving a frequency of $f_{closed} = \frac{3 V}{4 L}$. Because these two pipes produce the same frequency, we can set the expressions equal to each other:

$\frac{3 V}{2 l_0} = \frac{3 V}{4 L}$

By simplifying this equality, the common terms cancel out to leave $\frac{1}{2 l_0} = \frac{1}{4 L}$. Cross-multiplying results in $4 L = 2 l_0$, which confirms that the length of the open pipe is $2 L$.

An air column closed at one end resonates only at odd harmonics, where the length of the column satisfies the condition $L = (2n-1)\frac{\lambda}{4}$ for integers $n = 1, 2, 3...$. This structure implies that the first resonant length corresponds to the fundamental mode where $L = \frac{\lambda}{4}$, while the next larger length occurs at the third harmonic where $L = \frac{3\lambda}{4}$. Given that the smallest resonant length is 50 cm, we can determine the next value by noting that it is 3 times the fundamental length. Therefore, the next larger resonant length is 3 times 50 cm, which equals 150 cm.

The propagation speed of a transverse pulse on a string is determined by the expression $v = \sqrt{T/\mu}$, where $T$ is the tension and $\mu$ is the mass per unit length of the rope. Since the pulse frequency remains constant, the wavelength $\lambda$ is directly proportional to the square root of the tension, $\lambda \propto \sqrt{T}$. At the bottom end of the rope, the tension is created solely by the block of mass $m_2$, giving $T_1 = m_2g$, whereas at the top end, the rope supports the weight of both the block and its own mass $m_1$, resulting in a tension of $T_2 = (m_1 + m_2)g$. The ratio of the wavelength at the top to the wavelength at the bottom is $\sqrt{\frac{(m_1 + m_2)g}{m_2g}}$, which simplifies to $\sqrt{\frac{m_1 + m_2}{m_2}}$.

Closed organ pipes resonate with odd harmonics, where the fundamental frequency is given by $f = \frac{v}{4l_c}$, while open organ pipes support all integer harmonics, meaning the second overtone occurs at the third harmonic with a frequency of $f = \frac{3v}{2l_o}$. Equating these two expressions leads to $\frac{v}{4l_c} = \frac{3v}{2l_o}$. By canceling the speed of sound $v$ from both sides, the relationship simplifies to $\frac{1}{4l_c} = \frac{3}{2l_o}$, which reduces to $2l_o = 12l_c$ or $l_o = 6l_c$. Substituting the given length of the closed pipe, which is 20 cm, into this expression yields $l_o = 6 \times 20$ cm, resulting in a final length of 120 cm for the open organ pipe.

Resonant frequencies in a stretched string occur at integer multiples of the fundamental frequency, following the relationship $f_n = n f_0$. Because the frequencies are consecutive harmonics when no other resonant frequencies exist between them, the fundamental frequency is equivalent to the difference between any two adjacent observed frequencies. Subtracting the lower frequency from the higher one yields the fundamental frequency, $f_0 = 420 - 315 = 105$ Hz. This calculation confirms that the lowest possible resonant frequency for the string is 105 Hz.

The observed frequency of sound changes due to the Doppler effect, which depends only on the relative velocity component between the source and the observer along their line of sight. When a sound source moves at an angle to the line connecting it to a stationary observer, we must first determine this effective radial velocity by projecting the source's actual speed onto that line, which is $19.4 \cos(60^{\circ}) = 9.7$ m/s.

Since the source is approaching the observer, the waves are compressed, leading to an increase in the perceived frequency. We use the Doppler formula for a stationary observer and a moving source:

$f' = f_0 \left( \frac{v}{v - v_s} \right)$

Using the original frequency of 100 Hz, the speed of sound of 330 m/s, and the calculated radial speed of 9.7 m/s, the formula becomes $f' = 100 \left( \frac{330}{330 - 9.7} \right)$. Calculating this gives $f' = 100 \left( \frac{330}{320.3} \right)$, which simplifies to 103 Hz.

The adiabatic index, $\gamma$, connects the speed of sound in a gas to its temperature and molar mass through the equation $v = \sqrt{\frac{\gamma RT}{M}}$. Because the gas occupies 22.4 liters at NTP, 4.0 g of the substance corresponds to one mole, establishing the molar mass as $M = 0.004 \text{ kg mol}^{-1}$. Using the provided speed 952 m/s, temperature 273 K, and the gas constant $R = 8.3 \text{ J K}^{-1} \text{mol}^{-1}$, we solve for $\gamma = 1.6$, which is the ratio $8/5$. Given that $\gamma = \frac{C_P}{C_V}$ and $C_V = 5.0 \text{ J K}^{-1} \text{mol}^{-1}$, we calculate $C_P = \gamma C_V$, which results in a value of $C_P = 8.0 \text{ J K}^{-1} \text{mol}^{-1}$.

In a pipe closed at one end, natural oscillations occur as standing waves where the air column vibrates at specific harmonic frequencies. Because the pipe is closed at one end, the possible resonance modes are limited to the fundamental frequency and its odd-numbered harmonics.

The fundamental frequency is defined by the formula $f_0 = \frac{v}{4L}$. Substituting the given values where the length is 0.85 m and the velocity of sound is 340 m/s, the calculation results in $f_0 = \frac{340}{4 \times 0.85} = 100$ Hz. The sequence of possible oscillations consists of odd multiples of this fundamental frequency, calculated using the relationship $f_n = (2n+1)f_0$. The resulting frequencies are 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, and 1100 Hz. Since the next harmonic would be 1300 Hz, which exceeds the limit of 1250 Hz, there are 6 possible natural oscillations.

The Doppler effect dictates that the frequency heard by an observer changes when there is relative motion between the source of a sound and the observer. Because the motorcyclist is chasing the car, both are moving in the same direction, necessitating a formula that accounts for the observer approaching the source and the source receding from the observer. Converting the provided speeds yields an observer velocity of $v_o = 10$ m/s and a source velocity of $v_s = 5$ m/s, given a speed of sound of $v = 343$ m/s. Using the source frequency of $f = 1392$ Hz, the observed frequency $f'$ follows the relationship $f' = f \left( \frac{v + v_o}{v + v_s} \right)$, which gives $f' = 1392 \times \frac{343 + 10}{343 + 5}$. This expression simplifies to $1392 \times \frac{353}{348}$, resulting in an observed frequency of $1412$ Hz.

The Doppler effect dictates that the frequency observed by a listener depends on their motion relative to the source. Because the listener moves from source $P$ toward source $Q$, they simultaneously recede from $P$ and approach $Q$, causing a shift in the perceived sound frequencies from both sources. The apparent frequency from source $P$, from which the listener is moving away, is $f' = f \frac{v - v_L}{v}$, while the apparent frequency from source $Q$, toward which the listener is moving, is $f'' = f \frac{v + v_L}{v}$.

The beat frequency, representing the number of beats heard per second, is determined by the difference between these two shifted frequencies:
$\Delta f = f'' - f' = \frac{f(v + v_L)}{v} - \frac{f(v - v_L)}{v} = \frac{2fv_L}{v}$
By substituting the given values of 660 Hz for the source frequency $f$, 1 m/s for the listener speed $v_L$, and 330 m/s for the speed of sound $v$, the calculation simplifies to:
$\Delta f = \frac{2 \times 660 \times 1}{330} = 4$
This results in 4 beats heard per second.

The fundamental frequency of a sonometer wire is inversely proportional to its vibrating length, governed by the relationship $v \propto \frac{1}{L}$. When you divide the wire into three segments with frequencies in the ratio 1:3:5, the lengths of those segments must be in the inverse ratio of $1 : \frac{1}{3} : \frac{1}{5}$. Multiplying these terms by 15 to clear the denominators reveals that the lengths must be in the ratio 15:5:3.

For a wire with a total length of 100 cm, the sum of the ratio parts is $15 + 5 + 3 = 23$. This means the length of each unit part is $\frac{100}{23}$ cm. The first segment occupies 15 parts, making its length $\frac{1500}{23}$ cm, and the second segment occupies 5 parts, making its length $\frac{500}{23}$ cm. Therefore, the first bridge must be positioned at a distance of $\frac{1500}{23}$ cm from the end of the wire, and the second bridge must be placed at a distance of $\frac{1500}{23} + \frac{500}{23} = \frac{2000}{23}$ cm from the same end.

The general equation for a wave travelling in the positive x-direction is $y = A \sin(kx - \omega t)$, where $A$ denotes the amplitude, $k = 2\pi / \lambda$ is the angular wave number, and $\omega = 2\pi f$ is the angular frequency. Substituting the given wavelength $\lambda = 2\pi$ m and frequency $f = 1/\pi$ Hz, the angular wave number calculates as $k = 2\pi / 2\pi = 1$ rad/m, and the angular frequency becomes $\omega = 2\pi(1/\pi) = 2$ rad/s. With an amplitude of $A = 1$ m, the wave equation is $y = 1 \sin(1x - 2t)$, which simplifies directly to $y = \sin(x - 2t)$.

In a pipe that is open at both ends, the air molecules at the boundaries are free to oscillate with maximum displacement, creating antinodes at these locations. Because standing wave physics dictates that displacement antinodes align with pressure nodes, the amplitude of the pressure fluctuation is actually at a minimum at both ends, directly contradicting the notion that pressure changes are maximal there. This physical configuration allows the pipe to resonate at all integer multiples of the fundamental frequency, defined by $f_n = n f_1$, which ensures that both even and odd harmonics are present in the resulting sound.

Beats occur at a frequency defined by the absolute difference between two interfering sound sources, $b = |f_1 - f_2|$. With a known source frequency of 250 Hz and a beat frequency of 4 beats/s, the unknown frequency $f$ must satisfy the equation $|f - 250| = 4$, which leads to two possible candidate values: 254 Hz or 246 Hz. The second harmonic of the unknown source is $2f$, and the problem states that this harmonic produces 5 beats/s against a source of 513 Hz, implying $|2f - 513| = 5$. Substituting the first candidate value of 254 Hz into the expression for the second harmonic gives $2f = 508$, resulting in a beat frequency of $|508 - 513| = 5$, which matches the condition provided. Conversely, testing the second candidate value of 246 Hz gives a second harmonic of $2f = 492$, resulting in a beat frequency of $|492 - 513| = 21$, which is inconsistent with the problem statement. Consequently, the unknown frequency is 254 Hz.

When two sound waves with slightly different frequencies overlap, they interfere to create a periodic variation in loudness known as beats. The frequency of these beats is defined by the absolute difference between the frequencies of the individual sources, which can be extracted from the standard wave equation $y=A\sin(\omega t)$ by using the relationship $\omega = 2\pi f$. For the first wave $y_1=4\sin(600\pi t)$, the angular frequency is $600\pi$, which corresponds to a linear frequency of $300\text{ Hz}$. Similarly, for the second wave $y_2=5\sin(608\pi t)$, the angular frequency is $608\pi$, resulting in a linear frequency of $304\text{ Hz}$. Subtracting these two values yields a beat frequency of $4\text{ beats per second}$.

The ratio of maximum intensity to minimum intensity, commonly associated with the waxing and waning of the sound, depends on the amplitudes of the two waves. This ratio is calculated using the formula $\frac{I_{max}}{I_{min}}=\frac{(A_1+A_2)^2}{(A_1-A_2)^2}$. Given the amplitudes $A_1=4$ and $A_2=5$, the calculation becomes $\frac{(4+5)^2}{(4-5)^2}$, which simplifies to $\frac{9^2}{(-1)^2}$, resulting in a ratio of $81:1$. Consequently, the listener perceives a sound that fluctuates in intensity with this specific ratio at a rate of 4 times per second.

This problem is an application of the Doppler effect involving a sound wave emitted by a moving source that reflects off a stationary object and returns to that same moving source. Because the train is moving toward the stationary object, the sound initially emitted by the train is perceived at a higher frequency by the object, and when this sound reflects back, the train perceives the echo with a further frequency shift due to its own motion.

To calculate the frequency of the echo heard by the driver, we use the formula for a combined source and observer Doppler shift, which is expressed as $f' = f \left( \frac{v + v_T}{v - v_T} \right)$. Substituting the speed of sound of 330 m/s, the train speed of 220 m/s, and the source frequency of 1000 Hz, the calculation becomes $f' = 1000 \left( \frac{330 + 220}{330 - 220} \right)$. Simplifying the terms results in $1000 \times \frac{550}{110}$, which gives a final frequency of 5000 Hz.