Equation of a standing wave is expressed as $y = {\rm{ }}2Asin\omega tcoskx$. In this equation, quantity $\omega /k$ represents.

The transverse speed of particle of the string

WAVES MCQs for NEET UG PHYSICS — Medium, Page 2

31 Medium

Application +4 | -1

If length of stretched string is increased by 50% keeping mass per unit length same, then in order to keep its frequency of vibration constant, its tension must be increased by

View answer and explanation

Correct option: D

1.25

Option D (1.25) is correct because the tension must be increased by a factor of 2.25 to keep the frequency constant when the length is increased by 50%.

Solution approach

Correct Answer + Core Reason: 1.25. The tension must be increased by a factor of (1.5)² to keep the frequency constant.

Step-by-Step Solution:

Step 1: The frequency of vibration of a stretched string is given by $f = \frac{1}{{2L}}\sqrt{\frac{T}{\mu}}$ , where $L$ is the length, $T$ is the tension, and $\mu$ is the mass per unit length.

Step 2: We want to keep the frequency constant, so $f_1 = f_2$ .

Step 3: Let the initial length be $L_1 = L$ and the final length be $L_2 = 1.5L$ . Let the initial tension be $T_1 = T$ and the final tension be $T_2$ .

Step 4: Since the mass per unit length is constant, we have:
$\frac{1}{{2L_1}}\sqrt{\frac{T_1}{\mu}} = \frac{1}{{2L_2}}\sqrt{\frac{T_2}{\mu}}$
$\frac{1}{L}\sqrt{T} = \frac{1}{{1.5L}}\sqrt{T_2}$

Step 5: Square both sides: $\frac{1}{{{L^2}}}T = \frac{1}{{{{(1.5)}^2}{L^2}}}{T_2}$

Step 6: Solve for $T_2$ : $T_2 = {(1.5)^2}T = 2.25T$ .

Step 7: The increase in tension is $T_2 - T_1 = 2.25T - T = 1.25T$ . Therefore, the tension must be increased by a factor of 1.25.

Why other options are wrong:
- Option A (0.5): This is incorrect because it suggests a decrease in tension, which is not required to maintain constant frequency when the length is increased.
- Option B (1): This is incorrect because it implies the tension should be doubled, which is not the correct factor.
- Option C (0.75): This is incorrect because it implies a smaller increase in tension than is required.

Key Concepts:
- Frequency of a stretched string: $f = \frac{1}{{2L}}\sqrt{\frac{T}{\mu}}$

Memory Aids:
- Remember the relationship between frequency, length, and tension in a stretched string.

Quick Tips:
- When dealing with ratios, set up the equations carefully and simplify.

32 Medium

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Understanding +4 | -1

Which of the following equations represents a wave travelling along

y

-axis

View answer and explanation

Correct option: B

x = A\,\,\sin \left( {ky - \omega t} \right)

Option B is correct because the wave equation represents displacement along the x-axis as a function of y and t.

Solution approach

Correct Answer + Core Reason: $x = A\,\,\sin \left( {ky - \omega t} \right)$ . This equation represents a wave traveling along the y-axis.

Step-by-Step Solution:

Step 1: General Wave Equation: A wave traveling along the y-axis has the form $x = f(y, t)$ .

Step 2: Analysis of Option B: $x = A\,\,\sin \left( {ky - \omega t} \right)$ represents a wave where the displacement is along the x-axis and the wave propagates along the y-axis.

Why other options are wrong:
- Option A: $y = A\,\,\sin \left( {kx - \omega t} \right)$ represents a wave traveling along the x-axis.
- Option C: $y = A\,\,\sin \,\,ky\,\,\cos \,\,\omega t$ represents a standing wave along the y-axis.
- Option D: $y = A\,\,\cos \,\,ky\,\,\sin \,\,\omega t$ represents a standing wave along the y-axis.

Key Concepts:
- Wave Equation
- Wave Propagation
- Traveling Wave
- Standing Wave

Memory Aids:
- y-axis wave = x = f(y, t)

Quick Tips:
- Pay attention to which axis the displacement and propagation occur along.

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Application +4 | -1

A wire of density

9 \times {10^3}kg{m^{ - 3}}

is stretched between two clamps

1

m part and is subjected to an extension of

4.9 \times {10^{ - 4}}m

. The lowest frequency of transverse vibration in the wire is

Y = 9 \times {10^{10}}N{m^{ - 2}})

View answer and explanation

Correct option: B

35\mathrm{ }\,\,Hz

Option B (

35\mathrm{ }\,\,Hz

) is correct because it is the calculated fundamental frequency of the stretched wire using the given parameters.

Solution approach

Correct Answer + Core Reason: $35\mathrm{ }\,\,Hz$ . The fundamental frequency of the wire is calculated using the given density, extension, and Young's modulus.

Step-by-Step Solution:

Step 1: Identify the given parameters:\nDensity, $\rho = 9 \times {10^3} kg/m^3$ \nLength, $L = 1 m$ \nExtension, $\Delta L = 4.9 \times {10^{-4}} m$ \nYoung's modulus, $Y = 9 \times {10^{10}} N/m^2$

Step 2: Calculate the tension in the wire using Young's modulus:
$Y = \frac{Stress}{Strain} = \frac{T/A}{\Delta L/L}$ , where $T$ is the tension and $A$ is the area of cross-section.\nSo, $T = Y \times A \times \frac{\Delta L}{L}$

Step 3: Calculate the mass per unit length, $\mu$ :
$\mu = \frac{m}{L} = \rho \times A \times L / L = \rho A$

Step 4: Calculate the velocity of the transverse wave:
$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{Y \times A \times \frac{\Delta L}{L}}{\rho A}} = \sqrt{\frac{Y \times \Delta L}{\rho \times L}}$

Step 5: Substitute the given values:\nv = $\sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{9 \times 10^3 \times 1}} = \sqrt{\frac{9 \times 4.9 \times 10^6}{9 \times 10^3}} = \sqrt{4.9 \times 10^3} = \sqrt{4900} = 70 m/s$

Step 6: Calculate the fundamental frequency:
$f = \frac{v}{2L} = \frac{70}{2 \times 1} = 35 Hz$

Why other options are wrong:
- **Option A ( $40\mathrm{ }\,\,Hz$ ): This is incorrect due to a miscalculation in either the velocity or the frequency formula.
- Option C ( $30\mathrm{ }\,\,Hz$ ): This is incorrect due to a miscalculation in either the velocity or the frequency formula.
- Option D ( $25\mathrm{ }\,\,Hz$ ): This is incorrect due to a miscalculation in either the velocity or the frequency formula.

Key Concepts:
- Young's Modulus
- Wave velocity on a string
- Fundamental frequency

Memory Aids:**
- Remember the formula for wave velocity on a string: $v = \sqrt{\frac{T}{\mu}}$
- Remember the formula for fundamental frequency: $f = \frac{v}{2L}$

Quick Tips:
- Always check the units before substituting values into the formulas.
- Make sure to use the correct formula for the fundamental frequency.

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Remembering +4 | -1

Doppler phenomena is related with

View answer and explanation

Correct option: A

Pitch (frequency)

Option A (Pitch (frequency)) is correct because the Doppler effect describes the change in observed frequency (pitch) of a wave due to relative motion between the source and observer.

Solution approach

Correct Answer + Core Reason: Pitch (frequency). The Doppler effect is related to the change in observed frequency.

Step-by-Step Solution:

Step 1: The Doppler effect describes the change in frequency of a wave for an observer moving relative to the source of the wave.

Step 2: The pitch of a sound is directly related to its frequency.

Step 3: Therefore, the Doppler phenomenon is related to pitch (frequency).

Why other options are wrong:
- Option B: Loudness is incorrect; loudness is related to the amplitude of the sound wave.
- Option C: Quality is incorrect; quality is related to the timbre of the sound wave.
- Option D: Reflection is incorrect; reflection is a different phenomenon.

Key Concepts:
- Doppler effect
- Pitch and frequency

Memory Aids:
- Doppler effect changes the pitch.

Quick Tips:
- Remember the relationship between pitch and frequency.

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Understanding +4 | -1

Which of the following equations represents a wave?

View answer and explanation

Correct option: C

y = A\,\,\sin \left( {at - dx + c} \right)

Option C is correct because it represents a sinusoidal function with both time and position variables, which is characteristic of a wave.

Solution approach

Correct Answer + Core Reason: $y = A\,\,\sin \left( {at - dx + c} \right)$ . This equation represents a sinusoidal wave.

Step-by-Step Solution:

Step 1: General Wave Equation: A general wave equation has the form $y = f(x, t)$ , where $x$ is position and $t$ is time.

Step 2: Sinusoidal Wave: A sinusoidal wave is represented by a sine or cosine function with arguments involving both $x$ and $t$ .

Step 3: Analysis of Option C: $y = A\,\,\sin \left( {at - dx + c} \right)$ fits the description of a sinusoidal wave, where $a$ and $d$ are constants related to frequency and wave number, and $c$ is a phase constant.

Why other options are wrong:
- Option A: $y = A\sin \omega t$ represents simple harmonic motion at a fixed point, not a propagating wave.
- Option B: $y = A\,\,\cos \,kx$ represents a stationary wave or a snapshot of a wave at a particular time, not a propagating wave.
- Option D: $y = A\left( {\omega t - kx} \right)$ is a linear function, not a sinusoidal wave.

Key Concepts:
- Wave Equation
- Sinusoidal Wave
- Wave Propagation

Memory Aids:
- Wave = Sine/Cosine with x and t

Quick Tips:
- Look for sinusoidal functions with both position and time variables to identify wave equations.

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Remembering +4 | -1

The relation between frequency

n

, wavelength

\lambda

and velocity

v

of propagation of wave is

View answer and explanation

Correct option: C

n=\frac{v}{\lambda}

Option C (

n=\frac{v}{\lambda}

) is correct because it represents the fundamental relationship between frequency, wavelength, and velocity of a wave.

Solution approach

Correct Answer + Core Reason: $n=\frac{v}{\lambda}$ . This is the fundamental relationship between frequency, wavelength, and velocity of a wave.

Step-by-Step Solution:

Step 1: Recall the relationship between wave speed ( $v$ ), frequency ( $n$ ), and wavelength ( $\lambda$ ).

Step 2: The formula is $v = n \lambda$ .

Step 3: Rearrange the formula to solve for frequency ( $n$ ): $n = \frac{v}{\lambda}$ .

Step 4: The correct formula is $n = \frac{v}{\lambda}$ .

Why other options are wrong:
- Option A: $n=v \lambda$ is incorrect because it implies that the frequency is the product of velocity and wavelength, which is the opposite of the correct relationship.
- Option B: $n=\frac{\lambda}{v}$ is incorrect because it implies that the frequency is the ratio of wavelength to velocity, which is the inverse of the correct relationship.
- Option D: $n=\frac{1}{v}$ is incorrect because it only relates frequency to velocity and doesn't include the wavelength, which is necessary to describe a wave.

Key Concepts:
- Wave speed, frequency, and wavelength relationship

Memory Aids:
- Remember the formula: Velocity = Frequency x Wavelength (v = fλ)

Quick Tips:
- Always remember the units of each quantity: velocity (m/s), frequency (Hz), and wavelength (m).

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Understanding +4 | -1

When temperature increases, the frequency of a tuning fork

View answer and explanation

Correct option: B

Decreases

Option B (Decreases) is correct because as temperature increases, the speed of sound in the tuning fork decreases, leading to a decrease in frequency.

Solution approach

Correct Answer + Core Reason: The frequency of a tuning fork decreases when temperature increases.

Step-by-Step Solution:

Step 1: The frequency of a tuning fork depends on its physical properties, including its elasticity and density.

Step 2: As temperature increases, the elasticity of the tuning fork typically decreases, and its dimensions may slightly increase due to thermal expansion.

Step 3: These changes result in a decrease in the frequency of the tuning fork.

Why other options are wrong:
- Option A: The frequency decreases, not increases.
- Option C: The frequency changes with temperature.
- Option D: The frequency change depends on the material, but it generally decreases with increasing temperature.

Key Concepts:
- Frequency of a tuning fork
- Thermal expansion
- Elasticity

Memory Aids:
- Higher temperature, lower frequency (for tuning forks)

Quick Tips:
- Remember that temperature affects the physical properties of materials, which can affect their vibrational frequencies.

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Understanding +4 | -1

Equation of a standing wave is expressed as

y = \mathrm{ }2Asin\omega tcoskx

. In this equation, quantity

\omega /k

represents.

View answer and explanation

Correct option: B

The speed of either of component wave

Option B (The speed of either of component wave) is correct because in the equation of a standing wave,

\omega/k

represents the speed of the individual waves that interfere to create the standing wave.

Solution approach

Correct Answer + Core Reason: The speed of either of component wave. In a standing wave equation, $\omega/k$ represents the speed of the component waves.

Step-by-Step Solution:

Step 1: Recall the general form of a traveling wave: $y = A\sin(kx - \omega t)$ or $y = A\sin(kx + \omega t)$ .

Step 2: Identify the wave speed. The speed of the traveling wave is given by $v = \frac{{\omega}}{{k}}$ .

Step 3: Relate to the standing wave equation. The standing wave is formed by the superposition of two such waves traveling in opposite directions. Therefore, $\omega/k$ represents the speed of each of these component waves.

Why other options are wrong:
- Option A: A standing wave doesn't propagate; it's a stationary pattern. So, it doesn't have a speed in the same sense as a traveling wave.
- Option C: The transverse speed of a particle of the string is given by the time derivative of the displacement, which is not simply $\omega/k$ .
- Option D: $\omega/k$ clearly represents a physical quantity, the speed of the component waves.

Key Concepts:
- Traveling waves and their speed: $v = \frac{{\omega}}{{k}}$
- Standing waves as superposition of traveling waves

Memory Aids:
- Remember the relationship $v = \frac{{\omega}}{{k}}$ for traveling waves.

Quick Tips:
- Recognize that standing waves are formed by interfering traveling waves.

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Understanding +4 | -1

When two sound waves are superimposed, beats are produced when they have

View answer and explanation

Correct option: D

Different frequencies

Option D (Different frequencies) is correct because beats are produced due to the superposition of two waves with slightly different frequencies.

Solution approach

Correct Answer + Core Reason: Different frequencies. Beats are formed due to the interference of waves with slightly different frequencies.

Step-by-Step Solution:

Step 1: Beats are a result of the superposition of two sound waves with slightly different frequencies.

Step 2: When two waves with different frequencies interfere, they create alternating constructive and destructive interference, resulting in beats.

Step 3: The beat frequency is the absolute difference between the two frequencies.

Why other options are wrong:
- Option A: Different amplitudes and phase. While amplitude and phase differences can affect the interference pattern, they don't directly cause beats. Beats specifically require different frequencies.
- Option B: Different velocities. Different velocities would cause a change in wavelength, but not necessarily beats. Beats are a direct result of frequency differences.
- Option C: Different phases. A constant phase difference will cause interference, but not the periodic variation in amplitude that defines beats. Beats require a difference in frequencies.

Key Concepts:
- Superposition of waves
- Interference
- Beat frequency

Memory Aids:
- Beats require different frequencies.

Quick Tips:
- Remember that beats are a result of the difference in frequencies of two sound waves.

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Application +4 | -1

The length of a rubber cord is

{l_1}

when the work done on it is 4J and

{l_2}

when he work done on it is 9J. The length of the cord, when work done is 16J is

View answer and explanation

Correct option: D

2{l_2} - {l_1}

Option D (

2{l_2} - {l_1}

) is correct because it represents the length of the cord when the work done is 16J, derived from the relationship between work and extension.

Solution approach

Correct Answer + Core Reason: $2{l_2} - {l_1}$ . The length is determined by the relationship between work done and extension.

Step-by-Step Solution:

Step 1: Assume the length of the cord is $l_0$ when no work is done. The work done is related to the extension by $W = \frac{1}{2} k (l - l_0)^2$ , where $l$ is the length of the cord, and $k$ is a constant.

Step 2: We have three cases:\nCase 1: $4 = \frac{1}{2} k (l_1 - l_0)^2$ => $(l_1 - l_0)^2 = \frac{8}{k}$ \nCase 2: $9 = \frac{1}{2} k (l_2 - l_0)^2$ => $(l_2 - l_0)^2 = \frac{18}{k}$ \nCase 3: $16 = \frac{1}{2} k (l - l_0)^2$ => $(l - l_0)^2 = \frac{32}{k}$

Step 3: Taking the square root of each equation:
$l_1 - l_0 = \sqrt{\frac{8}{k}} = 2\sqrt{\frac{2}{k}}$
$l_2 - l_0 = \sqrt{\frac{18}{k}} = 3\sqrt{\frac{2}{k}}$
$l - l_0 = \sqrt{\frac{32}{k}} = 4\sqrt{\frac{2}{k}}$

Step 4: Let $x = \sqrt{\frac{2}{k}}$ . Then:
$l_1 - l_0 = 2x$
$l_2 - l_0 = 3x$
$l - l_0 = 4x$

Step 5: We want to express $l$ in terms of $l_1$ and $l_2$ . From the first two equations:
$l_1 = l_0 + 2x$
$l_2 = l_0 + 3x$ \nMultiply the first equation by 3 and the second by 2:
$3l_1 = 3l_0 + 6x$
$2l_2 = 2l_0 + 6x$ \nSubtract the second equation from the first:
$3l_1 - 2l_2 = l_0$

Step 6: Substitute $l_0 = 3l_1 - 2l_2$ into $l - l_0 = 4x$ :\nl = l_0 + 4x
$l = (3l_1 - 2l_2) + 4x$ \nSince $l_1 = l_0 + 2x$ , then $2x = l_1 - l_0 = l_1 - (3l_1 - 2l_2) = -2l_1 + 2l_2$ \nSo $4x = 2(-2l_1 + 2l_2) = -4l_1 + 4l_2$
$l = 3l_1 - 2l_2 - 4l_1 + 4l_2 = -l_1 + 2l_2 = 2l_2 - l_1$

Why other options are wrong:
- Option A: Incorrect due to incorrect manipulation of equations.
- Option B: Incorrect due to incorrect manipulation of equations.
- Option C: Incorrect due to incorrect manipulation of equations.

Key Concepts:
- Work done in stretching a cord
- Relationship between work and extension

Memory Aids:
- Remember the formula for work done in stretching a cord.

Quick Tips:
- Express all variables in terms of a common variable.

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Understanding +4 | -1

The minimum distance of reflector surface from the source for listening the echo of sound is

View answer and explanation

Correct option: D

16.5\,\,m

Option D (16.5 m) is correct because it represents the minimum distance required for the echo to be heard distinctly, considering the persistence of hearing.

Solution approach

Correct Answer + Core Reason: 16.5 m. The minimum distance is calculated based on the speed of sound and the persistence of hearing.

Step-by-Step Solution:

Step 1: Identify the given information and what needs to be found. We need to find the minimum distance for an echo to be heard.

Step 2: Recall the concept of persistence of hearing. The human ear can distinguish two sounds as separate only if the time interval between them is at least 0.1 seconds.

Step 3: Use the formula: Distance = Speed x Time. The sound travels from the source to the reflector and back to the source, so we need to consider twice the distance.

Step 4: Assume the speed of sound in air is approximately 330 m/s.

Step 5: Calculate the total distance the sound travels: Total distance = Speed x Time = 330 m/s x 0.1 s = 33 m.

Step 6: Since the sound travels to the reflector and back, the minimum distance to the reflector is half of the total distance: Minimum distance = Total distance / 2 = 33 m / 2 = 16.5 m.

Step 7: Final Answer: 16.5 m

Why other options are wrong:
- Option A (28 m): This distance is too large and doesn't correspond to the minimum distance required based on the persistence of hearing.
- Option B (18 m): This distance is also too large and doesn't account for the correct time interval and speed of sound.
- Option C (19 m): This distance is also too large and doesn't account for the correct time interval and speed of sound.

Key Concepts:
- Echo
- Speed of sound
- Persistence of hearing

Memory Aids:
- Remember that the sound travels twice the distance to the reflector.

Quick Tips:
- Always consider the persistence of hearing when calculating the minimum distance for an echo.

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Analyzing +4 | -1

Fundamental frequency of pipe is

100\,\,Hz

and other two frequencies are

300\,\,Hz

and

500\,\,Hz

, then

View answer and explanation

Correct option: C

One end is open and another end is closed

Option C (One end is open and another end is closed) is correct because the frequencies are odd multiples of the fundamental frequency.

Solution approach

Correct Answer + Core Reason: One end is open and another end is closed. The frequencies are odd multiples of the fundamental frequency.

Step-by-Step Solution:

Step 1: Identify the given frequencies: 100 Hz, 300 Hz, and 500 Hz.

Step 2: Observe the relationship between the frequencies. The frequencies are in the ratio 1:3:5, which are odd multiples of the fundamental frequency.

Step 3: Recall that for a pipe closed at one end, the frequencies are given by $f_n = (2n - 1)f_1$ , where $n = 1, 2, 3, ...$ and $f_1$ is the fundamental frequency.

Step 4: Conclude that one end is open and the other end is closed.

Why other options are wrong:
- Option A: Pipe is open at both the ends is incorrect. For an open pipe, the frequencies are integer multiples of the fundamental frequency (1:2:3:...).
- Option B: Pipe is closed at both the ends is incorrect. A pipe closed at both ends does not produce sound.
- Option D: None of the above is incorrect because option C is correct.

Key Concepts:
- Fundamental frequency
- Overtones
- Harmonics
- Organ pipes (open and closed)
- Frequency relationships in organ pipes

Memory Aids:
- Closed pipe: odd harmonics only.
- Open pipe: all harmonics.

Quick Tips:
- Remember the frequency relationships for open and closed pipes.

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Application +4 | -1

A plane wave is represented by

x = 1.2\,\,\sin \left( {314t + 12.56y} \right)

. Where

x

and

y

are distances measured along in

x

and

y

direction in meters and

t

is time in seconds. This wave has

View answer and explanation

Correct option: C

A wavelength of

0.5\,\,m

and travels in 

- ve\,\,y

direction

Option C (A wavelength of

0.5\mathrm{ }\,m

and travels in 

- ve\mathrm{ }\,y

direction) is correct because the wavelength is calculated from the wave number, and the direction is determined by the sign in the wave equation.

Solution approach

Correct Answer + Core Reason: A wavelength of $0.5\mathrm{ }\,m$ and travels in  $- ve\mathrm{ }\,y$ direction. The wavelength is calculated from the wave number, and the direction is determined by the sign in the wave equation.

Step-by-Step Solution:

Step 1: Identify the given information: The equation of the wave is $x = 1.2\,\sin(314t + 12.56y)$ .

Step 2: Recall the general form of a wave equation: $x = A\sin(\omega t + ky)$ .

Step 3: From the given equation, identify the angular frequency and wave number: $\omega = 314$ and $k = 12.56$ .

Step 4: Calculate the wavelength using the formula: $\lambda = \frac{2\pi}{k} = \frac{2\pi}{12.56} = \frac{2 \times 3.14}{12.56} = \frac{6.28}{12.56} = 0.5\mathrm{ }\,m$ .

Step 5: Determine the direction of propagation: Since the equation is of the form $x = A\sin(\omega t + ky)$ , the wave travels in the negative y-direction.

Step 6: Final Answer: The wavelength is $0.5\mathrm{ }\,m$ and the wave travels in the negative y-direction.

Why other options are wrong:
- **Option A (A wavelength of $0.25\mathrm{ }\,m$ and travels in $+ ve\mathrm{ }\,x$ direction): The wavelength is calculated incorrectly, and the direction is also wrong.
- Option B (A wavelength of $0.25\mathrm{ }\,m$ and travels in  $+ ve\mathrm{ }\,y$ direction): The wavelength is calculated incorrectly, and the direction is also wrong.
- Option D (A wavelength of $0.5\mathrm{ }\,m$ and travels in $- ve\mathrm{ }\,x$ direction): The wavelength is correct, but the direction is wrong. The wave travels along the y-axis, not the x-axis.

Key Concepts:
- Plane Wave Equation
- Wavelength
- Wave Number
- Direction of Propagation

Memory Aids:**
- Remember the relationship: $\lambda = \frac{2\pi}{k}$
- The sign between $\omega t$ and $ky$ determines the direction of propagation.

Quick Tips:
- Always identify the wave number and angular frequency correctly from the wave equation.
- Pay attention to the sign to determine the direction of propagation.

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Application +4 | -1

A wave travelling in positive

X

-direction with

A = 0.2\,m

has a velocity of

360\,\,m/sec.

. If

\lambda = 60\,\,m,

, then correct expression for the wave is

View answer and explanation

Correct option: C

y = 0.2\sin \left[ {2\pi \left( {6t - \frac{x}{{60}}} \right)} \right]

Option C (

y = 0.2\sin \left[ {2\pi \left( {6t - \frac{x}{{60}}} \right)} \right]

) is correct because it represents a wave traveling in the positive x-direction with the given amplitude, velocity, and wavelength.

Solution approach

Correct Answer + Core Reason: $y = 0.2\sin \left[ {2\pi \left( {6t - \frac{x}{{60}}} \right)} \right]$ . This equation represents a wave traveling in the positive x-direction with the given parameters.

Step-by-Step Solution:

Step 1: The general form of a wave traveling in the positive x-direction is $y = A\sin(\omega t - kx)$ .

Step 2: Given $A = 0.2\,\,m$ and $\lambda = 60\,\,m$ , we need to find $\omega$ and $k$ .

Step 3: Calculate the wave number $k = \frac{{2\pi}}{\lambda} = \frac{{2\pi}}{{60}} = \frac{\pi}{{30}}$ .

Step 4: Calculate the angular frequency $\omega = vk = 360 \times \frac{{\pi}}{{30}} = 12\pi$ .

Step 5: Substitute the values into the general equation: $y = 0.2\sin(12\pi t - \frac{{\pi}}{{30}}x)$ .

Step 6: Rewrite the equation: $y = 0.2\sin[2\pi(6t - \frac{x}{{60}})]$ .

Why other options are wrong:
- **Option A ( $y = 0.2\sin \left[ {2\pi \left( {6t + \frac{x}{{60}}} \right)} \right]$ ): Incorrect because it represents a wave traveling in the negative x-direction.
- Option B ( $y = 0.2\sin \left[ {\pi \left( {6t + \frac{x}{{60}}} \right)} \right]$ ): Incorrect because it represents a wave traveling in the negative x-direction and has incorrect angular frequency and wave number.
- Option D ( $y = 0.2\sin \left[ {\pi \left( {6t - \frac{x}{{60}}} \right)} \right]$ ): Incorrect because it has incorrect angular frequency and wave number.

Key Concepts:**
- General form of a wave: $y = A\sin(\omega t - kx)$
- Wave number: $k = \frac{{2\pi}}{\lambda}$
- Angular frequency: $\omega = vk$

Memory Aids:
- Remember the sign convention for waves traveling in the positive and negative x-directions.

Quick Tips:
- Always check the direction of propagation and the values of angular frequency and wave number.

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Understanding +4 | -1

Learned Indian classical vocalists do not like the accompaniment of a harmonium because

View answer and explanation

Correct option: D

Tempered scale is used in the harmonium

Option D is correct because a tempered scale in a harmonium has fixed intervals, unlike the just intonation preferred by classical vocalists.

Solution approach

Correct Answer + Core Reason: Tempered scale is used in the harmonium. This scale has fixed intervals that do not perfectly match the just intonation preferred by vocalists.

Step-by-Step Solution:

Step 1: Understand Tempered Scale: A tempered scale divides the octave into equal intervals, resulting in slightly out-of-tune intervals compared to just intonation.

Step 2: Understand Just Intonation: Just intonation aims for pure, consonant intervals based on simple ratios, which is preferred in classical vocal music.

Step 3: Harmonium and Tempered Scale: Harmoniums use a tempered scale, which clashes with the precise intonation sought by classical vocalists.

Why other options are wrong:
- Option A: Intensity is a matter of volume, not intonation preference.
- Option B: Shrillness is a timbre issue, not the core reason for disliking the harmonium's accompaniment.
- Option C: Diatonic scale is a type of scale, but the issue is the tempering of the intervals, not the scale type itself.

Key Concepts:
- Tempered Scale
- Just Intonation
- Musical Intervals

Memory Aids:
- Harmonium = Tempered = Not Pure

Quick Tips:
- Remember the difference between tempered and just intonation.

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Remembering +4 | -1

Sound travels in rocks in the form of:

View answer and explanation

Correct option: C

Both longitudinal and transverse elastic waves

Option C (Both longitudinal and transverse elastic waves) is correct because solids can support both types of waves due to their structure.

Solution approach

Correct Answer + Core Reason: Both longitudinal and transverse elastic waves. Rocks, being solids, can support both types of waves.

Step-by-Step Solution:

Step 1: Recall the types of waves that can travel through different media.

Step 2: Longitudinal waves (compressional waves) can travel through solids, liquids, and gases.

Step 3: Transverse waves (shear waves) can only travel through solids because they require a medium with shear strength.

Step 4: Since rocks are solids, they can support both longitudinal and transverse waves.

Why other options are wrong:
- Option A: Longitudinal elastic waves only is incorrect because rocks can also support transverse waves.
- Option B: Transverse elastic waves only is incorrect because rocks can also support longitudinal waves.
- Option D: Non-elastic waves is incorrect because sound waves are elastic waves.

Key Concepts:
- Longitudinal waves
- Transverse waves
- Wave propagation in different media

Memory Aids:
- Solids support both longitudinal and transverse waves.

Quick Tips:
- Remember that transverse waves require a medium with shear strength, which only solids possess.

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Understanding +4 | -1

Sound waves in air always longitudinal because

View answer and explanation

Correct option: B

Air does not have a modulus of rigidity

Option B (Air does not have a modulus of rigidity) is correct because longitudinal waves require a medium that can support compression and rarefaction, while transverse waves require a medium with rigidity.

Solution approach

Correct Answer + Core Reason: Air does not have a modulus of rigidity. Longitudinal waves require a medium that can support compression and rarefaction, while transverse waves require rigidity.

Step-by-Step Solution:

Step 1: Define longitudinal and transverse waves. Longitudinal waves are waves in which the displacement of the medium is parallel to the direction of propagation. Transverse waves are waves in which the displacement of the medium is perpendicular to the direction of propagation.

Step 2: Explain the requirement for a medium. Longitudinal waves can propagate through any medium that can be compressed, while transverse waves require a medium with shear strength (rigidity).

Step 3: Analyze air. Air is a gas and does not have a modulus of rigidity (shear strength). Therefore, it can only support longitudinal waves.

Why other options are wrong:
- Option A: While sound waves in air are inherently longitudinal, the reason is due to the properties of air, not just the nature of sound waves in general.
- Option C: The composition of air as a mixture of gases doesn't directly determine whether it supports longitudinal or transverse waves. The key factor is the lack of rigidity.
- Option D: The density of air is related to the speed of sound, but not directly to whether it can support longitudinal or transverse waves. The key factor is the lack of rigidity.

Key Concepts:
- Longitudinal waves
- Transverse waves
- Modulus of rigidity

Memory Aids:
- Gases and liquids can only support longitudinal waves.

Quick Tips:
- Remember that solids can support both longitudinal and transverse waves, while fluids (liquids and gases) can only support longitudinal waves.

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Understanding +4 | -1

To raise the pitch of a stringed musical instrument the player can

View answer and explanation

Correct option: D

Both (2) and (3 )

Option D (Both (2) and (3 )) is correct because tightening the string increases tension and shortening the string decreases the wavelength, both of which increase the frequency (pitch).

Solution approach

Correct Answer + Core Reason: Tightening the string and shortening the string both increase the pitch.

Step-by-Step Solution:

Step 1: The frequency of a vibrating string is given by: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$ , where L is the length, T is the tension, and $\mu$ is the linear mass density.

Step 2: To raise the pitch (increase the frequency), you can either increase the tension (T) or decrease the length (L).

Step 3: Loosening the string decreases the tension, which decreases the frequency.

Why other options are wrong:
- Option A: Loosening the string decreases the pitch.
- Option B: Tightening the string increases the pitch.
- Option C: Shortening the string increases the pitch.

Key Concepts:
- Frequency of a vibrating string
- Tension
- Length

Memory Aids:
- Higher tension = higher pitch
- Shorter string = higher pitch

Quick Tips:
- Remember the formula for the frequency of a vibrating string.

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Application +4 | -1

A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by

y\left( {x,t} \right) = 0.5\sin \left( {\frac{{5\pi }}{4}x} \right)\cos \left( {200\pi t} \right)

. What is the speed of the travelling wave moving in the positive x direction?

View answer and explanation

Correct option: A

160 m/s

Option A (160 m/s) is correct because the speed of the traveling wave is calculated as the ratio of the angular frequency to the wave number, which are extracted from the given standing wave equation.

Solution approach

Correct Answer + Core Reason: 160 m/s. The speed of the traveling wave is given by $v = \frac{{\omega}}{{k}}$ .

Step-by-Step Solution:

Step 1: Identify $\omega$ and $k$ from the given equation: $y(x,t) = 0.5\sin(\frac{{5\pi}}{4}x)\cos(200\pi t)$ .

Step 2: From the equation, $k = \frac{{5\pi}}{4}$ and $\omega = 200\pi$ .

Step 3: Calculate the speed: $v = \frac{{\omega}}{{k}} = \frac{{200\pi}}{{\frac{{5\pi}}{4}}} = \frac{{200\pi \times 4}}{{5\pi}} = \frac{{800\pi}}{{5\pi}} = 160 \text{{ m/s}}$ .

Why other options are wrong:
- Option B: 90 m/s is incorrect. This is likely a miscalculation.
- Option C: 180 m/s is incorrect. This is likely a miscalculation.
- Option D: 120 m/s is incorrect. This is likely a miscalculation.

Key Concepts:
- Standing waves
- Wave speed: $v = \frac{{\omega}}{{k}}$

Memory Aids:
- Remember the relationship between wave speed, angular frequency, and wave number.

Quick Tips:
- Extract $\omega$ and $k$ carefully from the given equation.

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Application +4 | -1

The length of the wire between two ends of
a sonometer is 100 cm. What should be
the positions of two bridges below the wire
so that the three segments of the wire
have their fundamental frequencies in the
ratio 1 : 3 : 5

View answer and explanation

Correct option: D

\frac{{1500}}{{23}} \, cm,\,\frac{{2000}}{{23}} \, cm

Option D (

\frac{{1500}}{{23}} \, cm,\frac{{2000}}{{23}} \, cm

) is correct because it satisfies the condition where the fundamental frequencies of the three segments are in the ratio 1:3:5.

Solution approach

Correct Answer + Core Reason: $\frac{{1500}}{{23}} \, cm,\frac{{2000}}{{23}} \, cm$ . The positions of the bridges are calculated based on the inverse relationship between frequency and length for a vibrating string.

Step-by-Step Solution:

Step 1: Define the lengths of the three segments as $l_1$ , $l_2$ , and $l_3$ . The total length is 100 cm, so $l_1 + l_2 + l_3 = 100$ .

Step 2: The fundamental frequencies are inversely proportional to the lengths of the segments. Given the frequency ratio 1:3:5, we have $f_1 : f_2 : f_3 = 1:3:5$ . Therefore, $l_1 : l_2 : l_3 = \frac{{1}}{{1}} : \frac{{1}}{{3}} : \frac{{1}}{{5}}$ .

Step 3: Let $l_1 = k$ , $l_2 = \frac{{k}}{{3}}$ , and $l_3 = \frac{{k}}{{5}}$ .

Step 4: Substitute these into the total length equation: $k + \frac{{k}}{{3}} + \frac{{k}}{{5}} = 100$ .

Step 5: Find a common denominator and solve for $k$ : $\frac{{15k + 5k + 3k}}{{15}} = 100$ , which simplifies to $\frac{{23k}}{{15}} = 100$ .

Step 6: Solve for $k$ : $k = \frac{{1500}}{{23}}$ .

Step 7: Calculate the lengths of the segments: $l_1 = \frac{{1500}}{{23}} \, cm$ , $l_2 = \frac{{1}}{{3}} \times \frac{{1500}}{{23}} = \frac{{500}}{{23}} \, cm$ , and $l_3 = \frac{{1}}{{5}} \times \frac{{1500}}{{23}} = \frac{{300}}{{23}} \, cm$ .

Step 8: The positions of the bridges are at $l_1$ and $l_1 + l_2$ from one end. Therefore, the positions are $\frac{{1500}}{{23}} \, cm$ and $\frac{{1500}}{{23}} + \frac{{500}}{{23}} = \frac{{2000}}{{23}} \, cm$ .

Final Answer: The positions of the two bridges are $\frac{{1500}}{{23}} \, cm$ and $\frac{{2000}}{{23}} \, cm$ .

Why other options are wrong:
- Option A: $\frac{{1500}}{{23}} \, cm,\frac{{500}}{{23}} \, cm$ : This option only gives the first length correctly. The second position is incorrect.
- Option B: $\frac{{1500}}{{23}} \, cm,\frac{{300}}{{23}} \, cm$ : This option only gives the first length correctly. The second position is incorrect.
- Option C: $\frac{{300}}{{23}} \, cm,\frac{{1500}}{{23}} \, cm$ : This option has the lengths reversed and the second position is incorrect.
- Option D: $\frac{{1500}}{{23}} \, cm,\frac{{2000}}{{23}} \, cm$ : This is the correct answer as shown in the solution.

Key Concepts:
- Fundamental frequency of a string
- Inverse relationship between frequency and length
- Ratios and proportions

Memory Aids:
- Frequency is inversely proportional to length (for a given tension and mass per unit length).

Quick Tips:
- Remember that the frequency is inversely proportional to the length of the string.
- When dealing with ratios, express the lengths in terms of a common variable.

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Analyzing +4 | -1

The equation of the propagating wave is

y = 25\,\,\,\sin \left( {20t + 5x} \right)

where

y

is displacement. Which of the following statements is not true

View answer and explanation

Correct option: B

The wave is propagating in positive

x

-direction

Option B (The wave is propagating in positive

x

-direction) is not true because the wave is propagating in the negative x-direction due to the positive sign between the time and position terms in the sine function.

Solution approach

Correct Answer + Core Reason: The wave is propagating in positive $x$ -direction. The sign between the time and position terms determines the direction of propagation.

Step-by-Step Solution:

Step 1: The given equation is $y = 25\sin(20t + 5x)$ .

Step 2: The general form of a wave equation is $y = A\sin(\omega t - kx)$ for a wave traveling in the positive x-direction and $y = A\sin(\omega t + kx)$ for a wave traveling in the negative x-direction.

Step 3: Comparing the given equation with the general form, we see that the sign between the $t$ and $x$ terms is positive, which means the wave is traveling in the negative x-direction.

Step 4: The amplitude of the wave is $A = 25$ units.

Step 5: The angular frequency is $\omega = 20$ rad/s, and the wave number is $k = 5$ rad/m.

Step 6: The velocity of the wave is $v = \frac{\omega}{k} = \frac{20}{5} = 4$ units.

Step 7: The maximum velocity of the particles is $v_{max} = A\omega = 25 \times 20 = 500$ units.

Why other options are wrong:
- **Option A (The amplitude of the wave is $25\mathrm{ }units$ ): This is true because the amplitude is directly given as 25 in the equation.
- Option C (The velocity of the wave is $4\mathrm{ }units$ ):** This is true because the velocity is calculated as $\frac{\omega}{k} = \frac{20}{5} = 4$ .
- **Option D (The maximum velocity of the particles is $500\mathrm{ }units$ ):** This is true because the maximum particle velocity is calculated as $A\omega = 25 \times 20 = 500$ .

Key Concepts:
- Wave equation
- Amplitude
- Angular frequency
- Wave number
- Wave velocity
- Particle velocity

Memory Aids:
- Remember the general form of a wave equation: $y = A\sin(\omega t - kx)$ or $y = A\sin(\omega t + kx)$ .

Quick Tips:
- Pay attention to the sign between the time and position terms to determine the direction of propagation.

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Analyzing +4 | -1

A tuning fork of frequency

480\,\,Hz

produces 

\mathrm{10}{\mkern 1mu} {\mkern 1mu} \,\,\mathrm{beats}\,\,\mathrm{per}\,\,\mathrm{sec}

when sounded with a vibrating sonometer string. What must have been the frequency of string if slight increase in tension produces few beats

{\mathrm{s}^{ - 1}}

than before?

View answer and explanation

Correct option: B

\begin{array}{llllllllllllllllllll} {470\,\,Hz} \end{array}\,

Option B (470 Hz) is correct because the frequency of the string must be lower than the tuning fork's frequency, and increasing tension reduces the beat frequency.

Solution approach

Correct Answer + Core Reason: 470 Hz. The frequency of the string is less than that of the tuning fork, and increasing tension reduces the number of beats.

Step-by-Step Solution:

Step 1: Identify the given information: Tuning fork frequency ( $f_1 = 480$ Hz), beat frequency = 10 Hz. Let the frequency of the string be $f_2$ .

Step 2: The beat frequency is given by the absolute difference between the two frequencies: $|f_1 - f_2| = 10$ Hz.

Step 3: This gives two possibilities for $f_2$ : $f_2 = 480 + 10 = 490$ Hz or $f_2 = 480 - 10 = 470$ Hz.

Step 4: The problem states that a slight increase in tension produces fewer beats. The frequency of a string is given by $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ , where $T$ is the tension, $L$ is the length, and $\mu$ is the linear mass density. Thus, increasing the tension increases the frequency of the string.

Step 5: If $f_2 = 490$ Hz, increasing the tension would further increase $f_2$ , increasing the difference $|f_1 - f_2|$ and thus increasing the beat frequency. This contradicts the problem statement.

Step 6: If $f_2 = 470$ Hz, increasing the tension would increase $f_2$ , reducing the difference $|f_1 - f_2|$ and thus decreasing the beat frequency. This agrees with the problem statement.

Step 7: Therefore, the frequency of the string must be 470 Hz.

Why other options are wrong:
- Option A (490 Hz): If the string's frequency was 490 Hz, increasing the tension would increase the beat frequency, contradicting the problem statement.
- Option C (460 Hz): If the string's frequency was 460 Hz, the initial beat frequency would be |480 - 460| = 20 Hz, not 10 Hz as given.
- Option D (480 Hz): If the string's frequency was 480 Hz, the initial beat frequency would be 0 Hz, not 10 Hz as given.

Key Concepts:
- Beat frequency: $|f_1 - f_2|$
- Frequency of a string: $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Memory Aids:
- Beat frequency is the absolute difference between two frequencies.
- Increasing tension increases the frequency of a string.

Quick Tips:
- Always consider both possibilities when calculating beat frequencies.
- Remember the relationship between tension and frequency of a string.

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Analyzing +4 | -1

The displacement

y

of a particle is given by

y = 4\,\,{\cos ^{ 2}}\left( {\frac{t}{2}} \right)\,\,\sin \left( {1000t} \right)

. This expression may be considered to be a result of the superposition of how many simple harmonic motions?

View answer and explanation

Correct option: B

3

Option B (3) is correct because the given expression can be rewritten as a superposition of three simple harmonic motions using trigonometric identities.

Solution approach

Correct Answer + Core Reason: 3. The expression can be rewritten as a superposition of three simple harmonic motions.

Step-by-Step Solution:

Step 1: Identify the given expression: $y = 4 \cos^2(\frac{t}{2}) \sin(1000t)$ .

Step 2: Use the trigonometric identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ : $y = 4 \left(\frac{1 + \cos(t)}{2}\right) \sin(1000t) = 2(1 + \cos(t)) \sin(1000t)$ .

Step 3: Expand the expression: $y = 2\sin(1000t) + 2\cos(t)\sin(1000t)$ .

Step 4: Use the trigonometric identity $2\cos(A)\sin(B) = \sin(A + B) - \sin(A - B)$ : $y = 2\sin(1000t) + \sin(1001t) + \sin(999t)$ .

Step 5: Observe that the expression is now a superposition of three simple harmonic motions with frequencies 999, 1000, and 1001.

Why other options are wrong:
- Option A: 2 is incorrect. The initial expression requires further simplification to reveal the three SHMs.
- Option C: 4 is incorrect. The initial expression requires simplification, and there are only three SHMs.
- Option D: 5 is incorrect. The initial expression requires simplification, and there are only three SHMs.

Key Concepts:
- Superposition of simple harmonic motions
- Trigonometric identities
- Angular frequency

Memory Aids:
- Remember the trigonometric identities for simplifying expressions.

Quick Tips:
- Always try to simplify trigonometric expressions to identify the individual SHMs.

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Remembering +4 | -1

Frequency range of the audible sound is

View answer and explanation

Correct option: B

20 \mathrm{~Hz}-20 \mathrm{~kHz}

Option B (

20 \mathrm{~Hz}-20 \mathrm{~kHz}

) is correct because it represents the typical range of human hearing.

Solution approach

Correct Answer + Core Reason: The audible range is 20 Hz to 20 kHz.

Step-by-Step Solution:

Step 1: Define audible range: The range of frequencies that humans can typically hear.

Step 2: The lower limit of human hearing is approximately 20 Hz.

Step 3: The upper limit of human hearing is approximately 20 kHz.

Why other options are wrong:
- Option A: $0 \mathrm{~Hz}-30 \mathrm{~Hz}$ is too low. Humans can't hear frequencies below 20 Hz.
- Option C: $20 \mathrm{~kHz}-20000 \mathrm{~kHz}$ is too high. The upper limit is around 20 kHz.
- Option D: $20 \mathrm{~kHz}-20 \mathrm{~MHz}$ is too high. The upper limit is around 20 kHz.

Key Concepts:
- Audible range
- Frequency

Memory Aids:
- Remember the approximate range: 20 Hz - 20 kHz

Quick Tips:
- Know the standard range of human hearing.

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Analyzing +4 | -1

The transverse displacement

y\left( {x,t} \right)

of a wave on a string is given by

y\left( {x,t} \right) = {e^{ - \left( {a{x^2} + b{t^2} + 2\sqrt {ab} xt} \right)}}

This represent a

View answer and explanation

Correct option: A

Wave moving in

x

- direction with speed

\sqrt {\frac{b}{a}}

Option A (Wave moving in

x

- direction with speed

\sqrt {\frac{b}{a}}

) is correct because the given equation represents a wave moving along the x-axis, and the speed can be determined from the coefficients of x and t.

Solution approach

Correct Answer + Core Reason: Wave moving in $x$ - direction with speed $\sqrt {\frac{b}{a}}$ . The equation represents a traveling wave, and its speed is determined by the ratio of the coefficients of $t$ and $x$ .

Step-by-Step Solution:

Step 1: Rewrite the given equation as $y(x,t) = e^{-((\sqrt{a}x + \sqrt{b}t)^2)}$ .

Step 2: This equation represents a traveling wave of the form $y = f(x + vt)$ or $y = f(x - vt)$ .

Step 3: Comparing the given equation with the general form, we can see that the wave is moving in the negative x-direction if the term inside the parenthesis is $(\sqrt{a}x + \sqrt{b}t)$ and in the positive x-direction if it's $(\sqrt{a}x - \sqrt{b}t)$ . Since we have $(\sqrt{a}x + \sqrt{b}t)$ , we can rewrite it as $(\sqrt{a}(x + \frac{\sqrt{b}}{\sqrt{a}}t))$ .

Step 4: The speed of the wave is given by the coefficient of $t$ divided by the coefficient of $x$ , which is $\frac{\sqrt{b}}{\sqrt{a}} = \sqrt{\frac{b}{a}}$ .

Step 5: Since the term is $(\sqrt{a}x + \sqrt{b}t)$ , the wave is moving in the negative x-direction. However, the question asks for the speed, which is a scalar quantity.

Why other options are wrong:
- **Option B (Standing wave of frequency $\sqrt b$ ):** This is incorrect because the equation represents a traveling wave, not a standing wave. Standing waves have the form $y = A\sin(kx)\cos(\omega t)$ .
- **Option C (Standing wave of frequency $\frac{1}{{\sqrt b }}$ ): This is incorrect because the equation represents a traveling wave, not a standing wave.
- Option D (Wave moving in $+ x$ direction with speed $\sqrt {\frac{a}{b}}$ ):** This is incorrect because the speed is $\sqrt{\frac{b}{a}}$ , not $\sqrt{\frac{a}{b}}$ , and the wave is moving in the negative x-direction.

Key Concepts:
- Traveling waves
- Wave speed
- Wave equation

Memory Aids:
- Remember the general form of a traveling wave: $y = f(x - vt)$ or $y = f(x + vt)$ .

Quick Tips:
- Rewrite the equation in the standard form to identify the speed and direction of the wave.

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Application +4 | -1

In a transverse progressive wave of amplitude A, the maximum particle velocity is four times its wave velocity. The wavelength of the wave is:-

View answer and explanation

Correct option: B

\frac{{\pi A}}{2}

Option B (

\frac{{\pi A}}{2}

) is correct because the wavelength is derived from the relationship between maximum particle velocity, wave velocity, and amplitude.

Solution approach

Correct Answer + Core Reason: $\frac{{\pi A}}{2}$ . The wavelength is calculated using the relationship between particle velocity, wave velocity, and amplitude.

Step-by-Step Solution:

Step 1: Define the maximum particle velocity: $v_{p,max} = A\omega$ .

Step 2: Define the wave velocity: $v = \frac{\omega}{k}$ , where $k = \frac{{2\pi}}{{\lambda}}$ .

Step 3: Given that $v_{p,max} = 4v$ , we have $A\omega = 4\frac{\omega}{k}$ .

Step 4: Simplify the equation: $A = \frac{4}{k}$ .

Step 5: Substitute $k = \frac{{2\pi}}{{\lambda}}$ : $A = \frac{4}{{\frac{{2\pi}}{{\lambda}}}} = \frac{{4\lambda}}{{2\pi}}$ .

Step 6: Solve for $\lambda$ : $\lambda = \frac{{2\pi A}}{4} = \frac{{\pi A}}{2}$ .

Why other options are wrong:
- **Option A ( $\pi A$ ):** This is incorrect because it would imply $v_{p,max} = 2v$ , not $4v$ .
- **Option C ( $\frac{{\pi A}}{4}$ ):** This is incorrect because it would imply $v_{p,max} = 8v$ , not $4v$ .
- **Option D ( $2\pi A$ ):** This is incorrect because it would imply $v_{p,max} = \frac{v}{2}$ , not $4v$ .

Key Concepts:
- Maximum particle velocity: $v_{p,max} = A\omega$
- Wave velocity: $v = \frac{\omega}{k}$
- Wave number: $k = \frac{{2\pi}}{{\lambda}}$

Memory Aids:
- Remember the relationship between particle velocity, wave velocity, amplitude, and wave number.

Quick Tips:
- Relate the given condition to the standard formulas and solve for the required parameter.

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Application +4 | -1

When a sound wave of wavelength

\lambda

is propagating in a medium, the maximum velocity of the particle is equal to the velocity. The amplitude of wave is

View answer and explanation

Correct option: C

\frac{\mathrm{\lambda }}{{2\mathrm{\pi }}}

Option C (

\frac{\mathrm{\lambda }}{{2\mathrm{\pi }}}

) is correct because the amplitude is related to the wavelength when the maximum particle velocity equals the wave velocity.

Solution approach

Correct Answer + Core Reason: $\frac{\mathrm{\lambda }}{{2\mathrm{\pi }}}$ . The amplitude is derived by equating the maximum particle velocity to the wave velocity.

Step-by-Step Solution:

Step 1: The equation of a sound wave is $y = A\sin(\omega t - kx)$ , where $A$ is the amplitude, $\omega$ is the angular frequency, $k$ is the wave number, $t$ is time, and $x$ is position.

Step 2: The particle velocity is given by $v_p = \frac{\partial y}{\partial t} = A\omega \cos(\omega t - kx)$ .

Step 3: The maximum particle velocity is $v_{p,max} = A\omega$ .

Step 4: The wave velocity is $v = \frac{\omega}{k}$ .

Step 5: Given that the maximum particle velocity is equal to the wave velocity, we have $A\omega = \frac{\omega}{k}$ .

Step 6: Solving for the amplitude, $A = \frac{1}{k}$ .

Step 7: Since $k = \frac{2\pi}{\lambda}$ , we have $A = \frac{1}{\frac{2\pi}{\lambda}} = \frac{\lambda}{2\pi}$ .

Why other options are wrong:
- **Option A ( $\lambda$ ): This is incorrect because it does not account for the relationship between wave number and wavelength.
- Option B ( $\frac{\mathrm{\lambda }}{2}$ ):** This is incorrect because it does not account for the factor of $\pi$ in the relationship between wave number and wavelength.
- **Option D ( $\frac{\mathrm{\lambda }}{{4\mathrm{\pi }}}$ ): This is incorrect because it has an extra factor of 2 in the denominator.

Key Concepts:
- Wave equation
- Particle velocity
- Wave velocity
- Wave number

Memory Aids:**
- Remember the relationship between wave number and wavelength: $k = \frac{2\pi}{\lambda}$ .

Quick Tips:
- Always differentiate the wave equation with respect to time to find the particle velocity.

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Application +4 | -1

The equation of a simple harmonic progressive wave is given by

y = A\,\,sin(100 - 3x)

. find the distance between

2

particles having a phase difference of

\frac{\pi }{3}

.

View answer and explanation

Correct option: A

\frac{\mathrm{\pi }}{9}\mathrm{m}

Option A (

\frac{\mathrm{\pi }}{9}\rm{{m}}

) is correct because it represents the distance corresponding to the given phase difference, calculated using the wave number.

Solution approach

Correct Answer + Core Reason: $\frac{\mathrm{\pi }}{9}\mathrm{m}$ . The distance is calculated using the relationship between phase difference and wave number.

Step-by-Step Solution:

Step 1: Identify the given information: The equation of the wave is $y = A\,\sin(100t - 3x)$ , and the phase difference is $\Delta \phi = \frac{\pi}{3}$ .

Step 2: Recall the relationship between phase difference, wave number ( $k$ ), and distance ( $\Delta x$ ): $\Delta \phi = k \Delta x$ .

Step 3: From the given equation, identify the wave number: $k = 3$ .

Step 4: Substitute the values into the equation: $\frac{\pi}{3} = 3 \Delta x$ .

Step 5: Solve for $\Delta x$ : $\Delta x = \frac{\pi}{3 \times 3} = \frac{\pi}{9}$ .

Step 6: Final Answer: The distance between the two particles is $\frac{\pi}{9} \text{ m}$ .

Why other options are wrong:
- **Option B ( $\frac{\mathrm{\pi }}{{18}}\mathrm{m}$ ):** This is incorrect because it might be calculated by dividing the phase difference by twice the wave number, i.e., $\frac{\pi/3}{6} = \frac{\pi}{18}$ .
- **Option C ( $\frac{\mathrm{\pi }}{6}\mathrm{m}$ ): This is incorrect because it might be calculated by incorrectly dividing the phase difference by the wave number squared, or some other incorrect manipulation of the formula.
- Option D ( $\frac{\mathrm{\pi }}{3}\mathrm{m}$ ): This is incorrect because it might be calculated by simply taking the phase difference as the distance, without considering the wave number.

Key Concepts:
- Simple Harmonic Progressive Wave
- Phase Difference
- Wave Number

Memory Aids:**
- Remember the relationship: $\Delta \phi = k \Delta x$

Quick Tips:
- Always identify the wave number correctly from the wave equation.
- Ensure the correct formula is used relating phase difference, wave number, and distance.

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Application +4 | -1

A sonometer is set on the floor of a lift. When the lift is at rest, the sonometer wire vibrates with fundamental frequency 256 Hz. When the lift goes up with acceleration

a = \frac{{9g}}{{16}}

vibration of the same wire changes to

View answer and explanation

Correct option: B

320 Hz

Option B (320 Hz) is correct because the frequency of the sonometer wire is proportional to the square root of the tension, and the effective weight increases with upward acceleration.

Solution approach

Correct Answer + Core Reason: 320 Hz. The frequency is proportional to the square root of the tension, which is affected by the lift's acceleration.

Step-by-Step Solution:

Step 1: Recall the formula for the fundamental frequency of a sonometer wire: $f = \frac{1}{2L}\sqrt{{\frac{T}{\mu}}}$ , where $f$ is the frequency, $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.

Step 2: The tension in the wire is proportional to the effective weight of the mass hanging from it. When the lift is at rest, the tension is $T = mg$ . When the lift is accelerating upwards with acceleration $a$ , the effective weight is $m(g+a)$ .

Step 3: Calculate the effective acceleration: $a = \frac{{9g}}{{16}}$ , so the effective gravity is $g_{eff} = g + a = g + \frac{{9g}}{{16}} = \frac{{25g}}{{16}}$ .

Step 4: The new tension is $T' = m g_{eff} = m \frac{{25g}}{{16}}$ .

Step 5: The new frequency $f'$ is given by: $f' = \frac{1}{2L}\sqrt{{\frac{{T'}}{\mu}}} = \frac{1}{2L}\sqrt{{\frac{{m \frac{{25g}}{{16}}}}{\mu}}} = \frac{1}{2L}\sqrt{{\frac{{25}}{{16}}}}\sqrt{{\frac{{mg}}{\mu}}} = \frac{5}{4} f$ .

Step 6: Calculate the new frequency: $f' = \frac{5}{4} \times 256 \text{{ Hz}} = 320 \text{{ Hz}}$ .

Why other options are wrong:
- Option A: 512 Hz. This is incorrect due to calculation errors.
- Option C: 256 Hz. This is incorrect because the frequency changes due to the lift's acceleration.
- Option D: 204 Hz. This is incorrect due to calculation errors.

Key Concepts:
- Fundamental frequency of a sonometer wire
- Effective weight in an accelerating frame

Memory Aids:
- Remember that upward acceleration increases effective weight.

Quick Tips:
- Account for the effect of acceleration on tension.

60 Medium

60 of 159Question permalink

Analyzing +4 | -1

Two waves travelling in a medium in the x-direction are represented by ,

{y_1} = A\sin \left( {\alpha t - \beta x} \right)

and

{y_2} = A\cos \left( {\beta x + \alpha t - \pi /4} \right)

where

{y_1}\;\,\,and\,\,\mathrm{ }{y_2}

are the displacements of the particles of the medium, t is time, and 

\alpha \,and\,\beta

are constants. The two waves have different

View answer and explanation

Correct option: B

Directions of propagation

Option B (Directions of propagation) is correct because the sign difference between the time and position terms in the wave equations determines the direction of propagation.

Solution approach

Correct Answer + Core Reason: Directions of propagation. The sign of the terms in the argument of the trigonometric functions determines the direction.

Step-by-Step Solution:

Step 1: Analyze the first wave: ${y_1} = A\sin \left( {\alpha t - \beta x} \right)$ . This wave propagates in the positive x-direction.

Step 2: Analyze the second wave: ${y_2} = A\cos \left( {\beta x + \alpha t - \pi /4} \right) = A\cos \left( {\alpha t + \beta x - \pi /4} \right)$ . This wave propagates in the negative x-direction.

Step 3: Since the waves propagate in opposite directions, they have different directions of propagation.

Step 4: Final Answer: Directions of propagation

Why other options are wrong:
- Option A (Speeds): The speed of a wave is given by $v = \frac{\alpha}{\beta}$ . Both waves have the same α and β, so they have the same speed.
- Option C (Wavelenghts): The wavelength is related to β by $k = \frac{{2\pi}}{\lambda} = \beta$ , so $\lambda = \frac{{2\pi}}{\beta}$ . Both waves have the same β, so they have the same wavelength.
- Option D (Frequencies): The frequency is related to α by $\omega = 2\pi f = \alpha$ , so $f = \frac{\alpha}{{2\pi}}$ . Both waves have the same α, so they have the same frequency.

Key Concepts:
- Wave propagation
- Wave equation
- Direction of propagation

Memory Aids:
- Remember that the sign between the time and position terms in the wave equation determines the direction of propagation.

Quick Tips:
- If the signs are opposite, the wave propagates in the positive x-direction. If the signs are the same, the wave propagates in the negative x-direction.

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