Neither longitudinal nor transverse

WAVES MCQs for NEET UG PHYSICS — Medium, Page 4

91 Medium

Understanding +4 | -1

Two tuning forks of frequency

{n_1}

and

{n_2}

produces n beats per second. If

{n_2}

and n are known,

{n_1}

may be given by

View answer and explanation

Correct option: C

{\mathrm{n}_2} \pm \mathrm{n}

Option C (

{\mathrm{n}_2} \pm \mathrm{n}

) is correct because the beat frequency is the absolute difference between the two frequencies, so

{n_1}

can be either higher or lower than

{n_2}

by n.

Solution approach

Correct Answer + Core Reason: The frequency ${n_1}$ can be ${n_2} \pm n$ because the beat frequency is the absolute difference between the two frequencies.

Step-by-Step Solution:

Step 1: Define beat frequency: The beat frequency is the absolute difference between two frequencies: $n = |n_1 - n_2|$

Step 2: Solve for $n_1$ : $n_1 = n_2 \pm n$

Why other options are wrong:
- Option A: $\frac{{{\mathrm{n}_2}}}{\mathrm{n}} + {\mathrm{n}_2}$ is not the correct formula for beat frequency.
- Option B: ${\mathrm{n}_2}\mathrm{n}$ is not the correct formula for beat frequency.
- Option D: $\frac{{{\mathrm{n}_2}}}{\mathrm{n}} - {\mathrm{n}_2}$ is not the correct formula for beat frequency.

Key Concepts:
- Beat frequency

Memory Aids:
- Beat frequency is the absolute difference: $|n_1 - n_2|$

Quick Tips:
- Remember that the beat frequency is always a positive value.

92 Medium

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Application +4 | -1

A cylindrical tube, closed at one end, has a fundamental frequency f in air. The

\frac{1}{4}th

of the tube is filled with water. The fundamental frequency of the air column is now

View answer and explanation

Correct option: B

\frac{{4f}}{3}

Option B (

\frac{{4f}}{3}

) is correct because filling 1/4 of the tube with water reduces the effective length of the air column, increasing the fundamental frequency.

Solution approach

Correct Answer + Core Reason: $\frac{{4f}}{3}$ . Filling the tube with water changes the effective length and thus the frequency.

Step-by-Step Solution:

Step 1: The fundamental frequency of a cylindrical tube closed at one end is given by $f = \frac{{v}}{{4L}}$ , where $v$ is the speed of sound in air and $L$ is the length of the tube.

Step 2: When $\frac{1}{4}$ th of the tube is filled with water, the length of the air column becomes $L' = L - \frac{L}{4} = \frac{3L}{4}$ .

Step 3: The new fundamental frequency $f'$ is given by $f' = \frac{{v}}{{4L'}} = \frac{{v}}{{4(\frac{3L}{4})}} = \frac{{v}}{{3L}}$ .

Step 4: We can express $f'$ in terms of the original frequency $f$ :
$f' = \frac{{v}}{{3L}} = \frac{4}{3} \times \frac{{v}}{{4L}} = \frac{{4}}{3}f$

Why other options are wrong:
- Option A: Incorrect due to incorrect calculation of the new frequency.
- Option C: Incorrect due to incorrect calculation of the new frequency.
- Option D: Incorrect due to incorrect calculation of the new frequency.

Key Concepts:
- Fundamental frequency of a closed pipe
- Effect of changing the length of the air column

Memory Aids:
- Remember the formula for the fundamental frequency of a closed pipe.

Quick Tips:
- Calculate the new length of the air column accurately.

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Application +4 | -1

The instantaneous displacement of a simple harmonic oscillator is given by

= a\,\,\,\cos \left[ {\omega t + \frac{\pi }{4}} \right]

. Its speed will be maximum at the time

View answer and explanation

Correct option: D

\frac{\pi }{{4\omega }}

Option D (

\frac{\pi }{{4\omega }}

) is correct because the speed is maximum when the displacement is zero, which occurs at

t = \frac{\pi }{{4\omega }}

for the given equation.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{{4\omega }}$ . The speed is maximum when the displacement is zero.

Step-by-Step Solution:

Step 1: The displacement is given by $x = a\cos(\omega t + \frac{\pi}{4})$ .

Step 2: The velocity is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = -a\omega \sin(\omega t + \frac{\pi}{4})$ .

Step 3: The speed is maximum when the absolute value of velocity is maximum. This occurs when $|\sin(\omega t + \frac{\pi}{4})| = 1$ .

Step 4: Therefore, $\omega t + \frac{\pi}{4} = \frac{\pi}{2} + n\pi$ , where $n$ is an integer.

Step 5: For the first maximum speed, let $n = 0$ : $\omega t + \frac{\pi}{4} = \frac{\pi}{2}$ .

Step 6: Solve for $t$ : $\omega t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$ . So, $t = \frac{\pi}{4\omega}$ .

Why other options are wrong:
- **Option A ( $\frac{{2\pi }}{\omega }$ ): This is incorrect because it corresponds to a full period of the oscillation, not the time when the speed is maximum.
- Option B ( $\frac{\omega }{{2\pi }}$ ): This is incorrect because it represents the frequency, not the time when the speed is maximum.
- Option C ( $\frac{\omega }{\pi }$ ): This is incorrect because it does not satisfy the condition for maximum speed.

Key Concepts:
- Simple Harmonic Motion
- Displacement
- Velocity
- Angular Frequency

Memory Aids:
- In SHM, maximum speed occurs at the equilibrium position (zero displacement).

Quick Tips:**
- Remember that velocity is the derivative of displacement.
- The speed is maximum when the absolute value of the sine function is 1.

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Application +4 | -1

The equation of a simple harmonic wave is given by

y = 3\sin \frac{\pi }{2}\left( {50t - x} \right)

Where x and y are in metres and t is in seconds. The ratio of maximum particle velocity to the wave velocity is

View answer and explanation

Correct option: D

\frac{3}{2}\pi

Option D (

\frac{3}{2}\pi

) is correct because the ratio of maximum particle velocity to wave velocity is calculated using the given wave equation.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}\pi$ . The ratio of maximum particle velocity to wave velocity is calculated using the given wave equation.

Step-by-Step Solution:

Step 1: The given wave equation is $y = 3\sin \frac{\pi}{2}(50t - x)$ .

Step 2: Comparing with the standard form $y = A\sin(\omega t - kx)$ , we have $A = 3$ m, $\omega = \frac{50\pi}{2} = 25\pi$ rad/s, and $k = \frac{\pi}{2}$ rad/m.

Step 3: The maximum particle velocity is $v_{max} = A\omega = 3(25\pi) = 75\pi$ m/s.

Step 4: The wave velocity is $v = \frac{\omega}{k} = \frac{25\pi}{\frac{\pi}{2}} = 50$ m/s.

Step 5: The ratio of maximum particle velocity to wave velocity is $\frac{v_{max}}{v} = \frac{75\pi}{50} = \frac{3\pi}{2}$ .

Why other options are wrong:
- Option A: $3\pi$ is incorrect because it is not the correct ratio of the velocities.
- Option B: $\frac{2}{3}\pi$ is incorrect because it is the inverse of the approximate correct ratio.
- Option C: $2\pi$ is incorrect because it is not the correct ratio of the velocities.

Key Concepts:
- Simple harmonic wave equation
- Maximum particle velocity
- Wave velocity

Memory Aids:
- $v_{max} = A\omega$ , $v = \frac{\omega}{k}$

Quick Tips:
- Remember to correctly identify the amplitude, angular frequency, and wave number from the given wave equation.

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Analyzing +4 | -1

The displacement of particles in a string stretched in the

x

direction is represented by

y

. Among the following expressions for

y

, those describing wave motion are :

View answer and explanation

Correct option: A

\cos k x \sin \omega t

Option A (

\cos k x \sin \omega t

) is correct because it represents a valid solution to the wave equation.

Solution approach

Correct Answer + Core Reason: $\cos k x \sin \omega t$ . This expression represents a valid solution to the wave equation.

Step-by-Step Solution:

Step 1: Recall the general form of a wave equation: $y = f(x \pm vt)$ , where $v$ is the wave velocity.

Step 2: Analyze each option:
- Option A: $y = \cos kx \sin \omega t$ . This can be written as $y = \frac{1}{2} [\sin(\omega t + kx) + \sin(\omega t - kx)]$ , which represents the superposition of two waves traveling in opposite directions. This is a valid wave motion.
- Option B: $y = k^2 x^2 - \omega^2 t^2$ . This is a quadratic equation in $x$ and $t$ , and it does not represent a wave motion.
- Option C: $y = \cos^2(kx + \omega t)$ . This can be written as $y = \frac{1}{2} [1 + \cos(2(kx + \omega t))]$ . This represents a wave motion.
- Option D: $y = \cos(k^2 x^2 - \omega^2 t^2)$ . This expression does not represent a standard wave motion because the arguments of the cosine function are squared.

Step 3: Determine which options represent wave motion.\nOption A represents a wave motion.

Why other options are wrong:
- Option B: $k^2 x^2-\omega^2 t^2$ does not represent a wave motion as it is a quadratic equation, not a sinusoidal function of (x ± vt).
- Option C: $\cos ^2(k x+\omega t)$ represents a wave motion, but the question asks for *those* describing wave motion, implying there might be more than one correct answer. However, only option A is provided.
- Option D: $\cos \left(k^2 x^2-\omega^2 t^2\right)$ does not represent a standard wave motion because the arguments of the cosine function are squared.

Key Concepts:
- Wave equation
- Superposition of waves
- General form of a wave function

Memory Aids:
- Remember the general form of a wave function: $f(x \pm vt)$ .

Quick Tips:
- Check if the given expression can be written in the form $f(x \pm vt)$ to determine if it represents a wave motion.

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Remembering +4 | -1

A travelling wave reflected at an open boundary undergoes a phase change of

View answer and explanation

Correct option: B

0

Option B (0) is correct because a wave reflected at an open boundary undergoes no phase change.

Solution approach

Correct Answer + Core Reason: 0. A wave reflected at an open boundary undergoes no phase change.

Step-by-Step Solution:

Step 1: When a wave is reflected at an open boundary (e.g., a wave on a string reflected at a free end), the reflected wave has the same phase as the incident wave.

Step 2: This means there is no phase change upon reflection.

Why other options are wrong:
- Option A: $\pi$ . This is the phase change that occurs when a wave is reflected at a fixed boundary.
- Option C: $\frac{\pi}{2}$ . This is not a typical phase change for wave reflection.
- Option D: $\frac{\pi}{3}$ . This is not a typical phase change for wave reflection.

Key Concepts:
- Wave Reflection
- Open Boundary
- Fixed Boundary
- Phase Change

Memory Aids:
- Remember that a wave reflected at an open boundary has no phase change, while a wave reflected at a fixed boundary has a phase change of $\pi$ .

Quick Tips:
- Distinguish between reflection at open and fixed boundaries.

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Application +4 | -1

If two coherent waves are represented by

{y_1} = 4\sin \omega t

and

{y_2} = 3\sin \left( {\omega t + \pi /3} \right)

interfere at a point, the amplitude of resulting wave will be about

View answer and explanation

Correct option: B

6.1

Option B (6.1) is correct because it is the approximate amplitude calculated using the superposition principle for coherent waves.

Solution approach

Correct Answer + Core Reason: 6.1. The amplitude of the resulting wave is calculated using the superposition principle for coherent waves.

Step-by-Step Solution:

Step 1: Identify the given equations: $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin(\omega t + \frac{\pi}{3})$ .

Step 2: Recall the formula for the resultant amplitude of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ : $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$ .

Step 3: Substitute the given values: $A_1 = 4$ , $A_2 = 3$ , and $\phi = \frac{\pi}{3}$ .

Step 4: Calculate the resultant amplitude: $A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos(\frac{\pi}{3})} = \sqrt{16 + 9 + 24(\frac{1}{2})} = \sqrt{16 + 9 + 12} = \sqrt{37} \approx 6.08$ .

Step 5: Round to the nearest tenth: $A \approx 6.1$ .

Why other options are wrong:
- Option A: 7.2 is incorrect. It might be a result of adding the amplitudes directly or using an incorrect formula.
- Option C: 5 is incorrect. It might be a result of subtracting the amplitudes or using an incorrect formula.
- Option D: 12 is incorrect. It might be a result of adding the amplitudes and multiplying by some factor.

Key Concepts:
- Superposition of waves
- Resultant amplitude
- Phase difference

Memory Aids:
- Remember the formula: $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$ .

Quick Tips:
- Make sure to use the correct phase difference in the formula.
- Pay attention to the units and round to the appropriate number of significant figures.

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Analyzing +4 | -1

Assertion(A): Speed of sound in

\mathrm{O}_2

gas at temperature

T

is

v

. If temperature is increased to

2 T

, the molecular oxygen breaks into atomic oxygen. Then, speed of sound under these conditions will become

2 v

.

Reason(R): With increase in humidity in air, speed of sound increases.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.

Option B (If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion) is correct because both the assertion and the reason are true statements, but the reason does not explain the assertion.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. Both statements are true, but unrelated.

Step-by-Step Solution:

Step 1: Analyze Assertion(A): Speed of sound in $\mathrm{O}_2$ gas at temperature $T$ is $v$ . If temperature is increased to $2 T$ , the molecular oxygen breaks into atomic oxygen. Then, speed of sound under these conditions will become $2 v$ .

Step 2: The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$ , where $\gamma$ is the adiabatic index, $R$ is the gas constant, $T$ is the temperature, and $M$ is the molar mass.

Step 3: Initially, the gas is $\mathrm{O}_2$ , so $M = 32 \mathrm{g/mol}$ and $\gamma = 1.4$ . The speed of sound is $v = \sqrt{\frac{1.4RT}{32}}$ .

Step 4: After the temperature is increased to $2T$ , the oxygen breaks into atomic oxygen, so the gas becomes $\mathrm{O}$ , with $M = 16 \mathrm{g/mol}$ and $\gamma = 1.67$ . The speed of sound is $v' = \sqrt{\frac{1.67R(2T)}{16}} = \sqrt{\frac{3.34RT}{16}}$ .

Step 5: Calculate the ratio of the speeds: $\frac{{v'}}{v} = \sqrt{\frac{{\frac{{3.34RT}}{{16}}}}{{\frac{{1.4RT}}{{32}}}}} = \sqrt{\frac{{3.34 \times 32}}{{1.4 \times 16}}} = \sqrt{\frac{{3.34 \times 2}}{{1.4}}} = \sqrt{\frac{{6.68}}{{1.4}}} = \sqrt{4.77} \approx 2.18 \approx 2$ . Therefore, $v' \approx 2v$ . The assertion is correct.

Step 6: Analyze Reason(R): With increase in humidity in air, speed of sound increases. This statement is correct. Humidity decreases the density of air, which increases the speed of sound.

Step 7: Since both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion, the answer is B.

Step 8: Final Answer: If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.

Why other options are wrong:
- Option A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.): Reason is not the correct explanation.
- Option C (If Assertion is correct but Reason is incorrect.): Reason is correct.
- Option D (If Assertion is incorrect but Reason is correct.): Assertion is correct.

Key Concepts:
- Speed of sound in gases
- Adiabatic index
- Molar mass
- Humidity and speed of sound

Memory Aids:
- Remember the formula for the speed of sound in a gas: $v = \sqrt{\frac{\gamma RT}{M}}$ .

Quick Tips:
- Pay attention to the changes in molar mass and adiabatic index when the gas composition changes.

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Analyzing +4 | -1

Two tuning fork,

A

and

B

produce notes of frequencies

258\,\,Hz

and

262\,\,Hz

. An unknown note sounded with a produces certain beats. When the same note is sounded with

B

, the beat frequency gets doubled, the unknown frequency is

View answer and explanation

Correct option: B

254\mathrm{ }\,Hz

Option B (

254\mathrm{ }\,Hz

) is correct because the beat frequency with A is 4 Hz, and with B it's 8 Hz, satisfying the condition that the beat frequency with B is doubled.

Solution approach

Correct Answer + Core Reason: $254\mathrm{ }\,Hz$ . The unknown frequency is determined by analyzing the beat frequencies produced with two known frequencies.

Step-by-Step Solution:

Step 1: Let the unknown frequency be $f$ .

Step 2: The beat frequency with A is $|f - 258| = x$ , and the beat frequency with B is $|f - 262| = 2x$ .

Step 3: We have two cases:\nCase 1: $f < 258$ . Then $258 - f = x$ and $262 - f = 2x$ . Substituting $x = 258 - f$ into the second equation gives $262 - f = 2(258 - f) \Rightarrow 262 - f = 516 - 2f \Rightarrow f = 516 - 262 = 254 \mathrm{Hz}$ .\nCase 2: $f > 262$ . Then $f - 258 = x$ and $f - 262 = 2x$ . Substituting $x = f - 258$ into the second equation gives $f - 262 = 2(f - 258) \Rightarrow f - 262 = 2f - 516 \Rightarrow f = 516 - 262 = 254 \mathrm{Hz}$ . This contradicts the assumption that $f > 262$ .\nCase 3: $258 < f < 262$ . Then $f - 258 = x$ and $262 - f = 2x$ . Substituting $x = f - 258$ into the second equation gives $262 - f = 2(f - 258) \Rightarrow 262 - f = 2f - 516 \Rightarrow 3f = 778 \Rightarrow f = \frac{778}{3} \approx 259.33 \mathrm{Hz}$ . Then $x = 259.33 - 258 = 1.33$ and $2x = 2.66$ , but $262 - 259.33 = 2.67 \approx 2.66$ . This is a possible solution, but it's not an integer.

Step 4: From Case 1, $f = 254 \mathrm{Hz}$ . Then $x = |254 - 258| = 4$ and $2x = 8$ . Also, $|254 - 262| = 8$ , which satisfies the condition.

Step 5: Final Answer: $254\mathrm{ }\,Hz$

Why other options are wrong:
- Option A: $256\mathrm{ }\,Hz$ : If $f = 256$ , then $|256 - 258| = 2$ and $|256 - 262| = 6$ , which is not double.
- Option C: $300\mathrm{ }\,Hz$ : If $f = 300$ , then $|300 - 258| = 42$ and $|300 - 262| = 38$ , which is not double.
- Option D: $280\mathrm{ }\,Hz$ : If $f = 280$ , then $|280 - 258| = 22$ and $|280 - 262| = 18$ , which is not double.

Key Concepts:
- Beat frequency
- Superposition of waves

Memory Aids:
- Beat frequency is the absolute difference between two frequencies.

Quick Tips:
- Consider all possible cases when dealing with absolute values.

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Analyzing +4 | -1

A tuning fork

A

of frequency 512 Hz produce 5 beats/sec. When sounded with another tuning fork

B

of unknown frequency. If

B

is loaded with wax the numbers of beats is again 5 beats/second. The frequency of tuning fork

B

before it was loaded is:

View answer and explanation

Correct option: C

517 Hz

Option C (517 Hz) is correct because adding wax to tuning fork B decreases its frequency, and since the number of beats remained the same, the original frequency must have been higher than 512 Hz.

Solution approach

Correct Answer + Core Reason: 517 Hz. The frequency of tuning fork B must be higher than 512 Hz, and decreases when loaded with wax.

Step-by-Step Solution:

Step 1: Let $f_A = 512 \text{ Hz}$ be the frequency of tuning fork A and $f_B$ be the frequency of tuning fork B.

Step 2: The number of beats is $|f_A - f_B| = 5$ .

Step 3: This means $f_B$ could be either $507 \text{ Hz}$ or $517 \text{ Hz}$ .

Step 4: When tuning fork B is loaded with wax, its frequency decreases. Let the new frequency be $f_B'$ .

Step 5: The number of beats remains unchanged, so $|f_A - f_B'| = 5$ .

Step 6: If $f_B = 507 \text{ Hz}$ , then $f_B'$ would be even lower, and the number of beats would increase, which contradicts the given information. Therefore, $f_B$ cannot be $507 \text{ Hz}$ .

Step 7: If $f_B = 517 \text{ Hz}$ , then $f_B'$ would be lower, and the number of beats could remain the same. Therefore, $f_B = 517 \text{ Hz}$ .

Why other options are wrong:
- Option A: 502 Hz is incorrect; loading the tuning fork would increase the number of beats.
- Option B: 507 Hz is incorrect; loading the tuning fork would increase the number of beats.
- Option D: 522 Hz is incorrect; loading the tuning fork would change the number of beats.

Key Concepts:
- Beat frequency

Memory Aids:
- Loading a tuning fork decreases its frequency.

Quick Tips:
- Consider both possibilities for the initial frequency and eliminate based on the effect of loading.

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Understanding +4 | -1

Speed of sound at constant temperature depends on

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because the speed of sound at a constant temperature depends on the properties of the gas, but not directly on pressure or density individually. It depends on the ratio of pressure and density.

Solution approach

Correct Answer + Core Reason: None of the above. At constant temperature, the speed of sound depends on the properties of the gas, but not directly on pressure or density individually.

Step-by-Step Solution:

Step 1: Recall the formula for the speed of sound in a gas.\nThe speed of sound in a gas is given by $v = \sqrt{\frac{\gamma P}{\rho}}$ , where $\gamma$ is the adiabatic index, P is the pressure, and $\rho$ is the density.

Step 2: Apply the ideal gas law.\nFrom the ideal gas law, $PV = nRT$ , where n is the number of moles, R is the ideal gas constant, and T is the temperature. We can write $P = \frac{nRT}{V} = \frac{m}{M} \frac{RT}{V} = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}{M}$ , where m is the mass, M is the molar mass, and $\rho$ is the density.

Step 3: Substitute the ideal gas law into the speed of sound formula.
$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma (\rho \frac{RT}{M})}{\rho}} = \sqrt{\frac{\gamma RT}{M}}$

Step 4: Analyze the result.\nAt constant temperature, the speed of sound depends only on the gas constant R, the adiabatic index $\gamma$ , the temperature T, and the molar mass M. It does not depend on pressure or density independently.

Why other options are wrong:
- Option A: Pressure alone does not determine the speed of sound at constant temperature.
- Option B: Density alone does not determine the speed of sound at constant temperature.
- Option C: Neither pressure nor density alone determines the speed of sound at constant temperature.

Key Concepts:
- Speed of sound in a gas
- Ideal gas law
- Adiabatic index

Memory Aids:
- Remember the formula $v = \sqrt{\frac{\gamma RT}{M}}$

Quick Tips:
- Use the ideal gas law to relate pressure, density, and temperature.

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Analyzing +4 | -1

The wave function of a pulse is given by

y=\frac{3}{(2 x+3 t)^2}

where

x

and

y

are in metre and

t

is in second.

i) Identify the direction of propagation.

ii) Determine the wave velocity of the pulse.

View answer and explanation

Correct option: A

-1.5 \mathrm{~m} / \mathrm{s}

Option A (-1.5 m/s) is correct because the wave is propagating in the negative x-direction and the velocity is determined by the ratio of the coefficients of t and x.

Solution approach

Correct Answer + Core Reason: -1.5 m/s. The wave propagates in the negative x-direction with a velocity determined by the coefficients of x and t.

Step-by-Step Solution:

Step 1: Identify the general form of a wave function: $y = f(ax + bt)$ or $y = f(ax - bt)$ .

Step 2: Compare the given wave function $y = \frac{{3}}{{(2x + 3t)^2}}$ with the general form.

Step 3: Determine the direction of propagation. Since the term is $(2x + 3t)$ , it can be written as $2(x + \frac{{3}}{{2}}t)$ . This indicates propagation in the negative x-direction.

Step 4: Calculate the wave velocity. The wave velocity $v$ is given by $v = -\frac{{\text{{coefficient of }} t}}{{\text{{coefficient of }} x}} = -\frac{{3}}{{2}} = -1.5 \text{{ m/s}}$ .

Step 5: Final Answer: The wave velocity is -1.5 m/s.

Why other options are wrong:
- Option B: -2.5 m/s is incorrect. It is not the correct ratio of the coefficients of t and x.
- Option C: 1.5 m/s is incorrect. It has the correct magnitude but the wrong sign. The wave propagates in the negative x-direction.
- Option D: 2.5 m/s is incorrect. It has the wrong magnitude and sign. The wave propagates in the negative x-direction, and the velocity is not 2.5 m/s.

Key Concepts:
- Wave propagation
- Wave velocity
- General form of a wave function

Memory Aids:
- If the argument of the wave function is of the form (x + vt), the wave propagates in the negative x-direction. If it is of the form (x - vt), the wave propagates in the positive x-direction.

Quick Tips:
- Always check the sign to determine the direction of propagation.
- The wave velocity is the ratio of the coefficient of time to the coefficient of position.

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Analyzing +4 | -1

The phase difference between the two given waves is

{y_1} = {a_1}\sin \left( {\omega t - \frac{{2\pi }}{\lambda }X} \right)\,\,\,{y_2} = {a_2}\cos \left( {\omega t - \frac{{2\pi }}{\lambda }X + \phi } \right)\,

View answer and explanation

Correct option: C

\frac{\pi }{2} + \left( \phi \right)

Option C (

\frac{\pi }{2} + \left( \phi \right)

) is correct because the phase difference accounts for the cosine function and the additional phase angle

\phi

.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{2} + \left( \phi \right)$ . The phase difference includes the π/2 shift from cosine and the additional phase φ.

Step-by-Step Solution:

Step 1: Rewrite the second wave as a sine function: ${y_2} = {a_2}\cos \left( {\omega t - \frac{{2\pi }}{\lambda }X + \phi } \right) = {a_2}\sin \left( {\omega t - \frac{{2\pi }}{\lambda }X + \phi + \frac{\pi }{2}} \right)$ .

Step 2: Compare the phases of the two waves: ${y_1} = {a_1}\sin \left( {\omega t - \frac{{2\pi }}{\lambda }X} \right)$ and ${y_2} = {a_2}\sin \left( {\omega t - \frac{{2\pi }}{\lambda }X + \phi + \frac{\pi }{2}} \right)$ .

Step 3: The phase difference is the difference between the phases: $\Delta \varphi = \left( {\omega t - \frac{{2\pi }}{\lambda }X + \phi + \frac{\pi }{2}} \right) - \left( {\omega t - \frac{{2\pi }}{\lambda }X} \right) = \phi + \frac{\pi }{2}$ .

Step 4: Final Answer: $\frac{\pi }{2} + \left( \phi \right)$

Why other options are wrong:
- **Option A ( $\frac{{2\pi }}{\lambda }\left( \phi \right)$ ): This is incorrect because it involves the wavelength and the phase φ, which is not the correct phase difference.
- Option B ( $\phi$ ): This is incorrect because it doesn't account for the π/2 phase shift due to the cosine function.
- Option D ( $\left( \phi \right) - \frac{\pi }{2}$ ): This is incorrect because it subtracts π/2 instead of adding it.

Key Concepts:
- Phase difference
- Trigonometric identities
- Wave equation

Memory Aids:**
- Remember that $\cos(x) = \sin(x + \frac{\pi}{2})$ .

Quick Tips:
- Always convert cosine functions to sine functions before calculating the phase difference.

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Analyzing +4 | -1

A source of sound gives

5

beats

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

when sounded with another source of frequency

100\,\,Hz

. The second harmonic of the source together with a source of frequency

205\,\,Hz

gives

5\,\,beats\,\,{s^{( - 1)}}

. What is the frequency of the source?

View answer and explanation

Correct option: A

105\mathrm{ }Hz

Option A (

105\mathrm{ }Hz

) is correct because it satisfies both beat frequency conditions: 5 Hz with 100 Hz and 5 Hz with 205 Hz for the second harmonic.

Solution approach

Correct Answer + Core Reason: $105\mathrm{ }Hz$ . The frequency of the source is determined by analyzing the beat frequencies produced with known frequencies and its second harmonic.

Step-by-Step Solution:

Step 1: Let the frequency of the source be $f$ .

Step 2: The beat frequency with 100 Hz is $|f - 100| = 5$ . This gives two possible values for $f$ : $f = 105 \mathrm{Hz}$ or $f = 95 \mathrm{Hz}$ .

Step 3: The second harmonic of the source is $2f$ . The beat frequency of the second harmonic with 205 Hz is $|2f - 205| = 5$ . This gives two possible values for $2f$ : $2f = 210 \mathrm{Hz}$ or $2f = 200 \mathrm{Hz}$ .\nSo, $f = 105 \mathrm{Hz}$ or $f = 100 \mathrm{Hz}$ .

Step 4: Comparing the possible values of $f$ from Step 2 and Step 3, we see that $f = 105 \mathrm{Hz}$ is a possible solution.\nIf $f = 105 \mathrm{Hz}$ , then $|105 - 100| = 5$ and $|2(105) - 205| = |210 - 205| = 5$ . This satisfies both conditions.\nIf $f = 95 \mathrm{Hz}$ , then $|95 - 100| = 5$ and $|2(95) - 205| = |190 - 205| = 15$ , which does not satisfy the second condition.

Step 5: Final Answer: $105\mathrm{ }Hz$

Why other options are wrong:
- Option B: $205\mathrm{ }Hz$ : This is incorrect as it doesn't satisfy the first beat condition.
- Option C: $95\mathrm{ }Hz$ : This is incorrect as the second harmonic doesn't produce 5 beats with 205 Hz.
- Option D: $100\mathrm{ }Hz$ : This is incorrect as it doesn't satisfy the second beat condition.

Key Concepts:
- Beat frequency
- Harmonics

Memory Aids:
- Beat frequency is the absolute difference between two frequencies.

Quick Tips:
- Consider all possible cases when dealing with absolute values.
- Check if the solution satisfies all given conditions.

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Analyzing +4 | -1

Assertion: For same length but different gas COP fundamental frequency are different.
Reason: Here frequency is proportional to velocity for same length given.

View answer and explanation

Correct option: A

If both Assertion & reason are True & the Reason is a correct explanation of the Assertion

Option A is correct because the fundamental frequency depends on the velocity of sound in the gas, which is different for different gases, and the frequency is directly proportional to the velocity for a given length.

Solution approach

Correct Answer + Core Reason: If both Assertion & reason are True & the Reason is a correct explanation of the Assertion. The fundamental frequency depends on the gas and is proportional to velocity.

Step-by-Step Solution:

Step 1: Recall the formula for the fundamental frequency of a gas in a tube of length $L$ : $f = \frac{v}{2L}$ , where $v$ is the speed of sound in the gas.

Step 2: Analyze the Assertion: For the same length $L$ , but different gases, the fundamental frequency $f$ will be different because the speed of sound $v$ is different for different gases.

Step 3: Analyze the Reason: The frequency $f$ is directly proportional to the velocity $v$ for a given length $L$ . This is evident from the formula $f = \frac{v}{2L}$ .

Step 4: Determine if the Reason is a correct explanation of the Assertion: Since the fundamental frequency depends on the speed of sound in the gas, and the speed of sound is different for different gases, the fundamental frequency will be different for different gases in the same length tube. The Reason correctly explains why the Assertion is true.

Why other options are wrong:
- Option B: The Reason *is* a correct explanation of the Assertion.
- Option C: The Reason is True.
- Option D: Both Assertion and Reason are True.

Key Concepts:
- Fundamental frequency
- Speed of sound in a gas
- Relationship between frequency, velocity, and length: $f = \frac{v}{2L}$

Memory Aids:
- Remember the formula for the fundamental frequency: $f = \frac{v}{2L}$ .
- Recall that the speed of sound depends on the gas.

Quick Tips:
- Understand the relationship between frequency, velocity, and length in wave phenomena.
- Remember that different gases have different speeds of sound.

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Application +4 | -1

On the superposition of two harmonic oscillations represented by

{X_1} = a\sin \left( {\omega t + {\phi _1}} \right)

and

{X_2} = a\sin \left( {\omega t + {\phi _2}} \right)

a resulting oscillation with the same time period and same amplitude is obtained. The value of

{\phi _1} - {\phi _2}

is

View answer and explanation

Correct option: A

120\,^\circ C

Option A (

120\,^\circ C

) is correct because when two oscillations of the same amplitude result in an oscillation of the same amplitude, the phase difference must be 120 degrees.

Solution approach

Correct Answer + Core Reason: $120\\,^\circ C$ . The phase difference is such that the resultant amplitude is the same as the individual amplitudes.

Step-by-Step Solution:

Step 1: Recall the formula for the resultant amplitude of two waves with the same frequency: $A^2 = A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi_1 - \phi_2)$ , where $A$ is the resultant amplitude, $A_1$ and $A_2$ are the individual amplitudes, and $(\phi_1 - \phi_2)$ is the phase difference.

Step 2: In this case, $A_1 = A_2 = a$ and $A = a$ . Substitute these values into the formula:
$a^2 = a^2 + a^2 + 2a^2 \cos(\phi_1 - \phi_2)$

Step 3: Simplify the equation:
$a^2 = 2a^2 + 2a^2 \cos(\phi_1 - \phi_2)$
$-a^2 = 2a^2 \cos(\phi_1 - \phi_2)$
\cos(φ_1 - φ_2) = - $\frac{1}{2}$ $\n\n**Step 4:** Solve for the phase difference:\n\phi_1 - \phi_2 = \arccos(-\frac{1}{2}) = 120^\circ$

Why other options are wrong:
- Option B: $90\\,^\circ C$ . This is incorrect because it would result in a different resultant amplitude.
- Option C: $60\\,^\circ C$ . This is incorrect because it would result in a different resultant amplitude.
- Option D: $15\\,^\circ C$ . This is incorrect because it would result in a different resultant amplitude.

Key Concepts:
- Superposition of waves
- Resultant amplitude
- Phase difference

Memory Aids:
- Remember the formula for the resultant amplitude.

Quick Tips:
- Visualize the vector addition of the amplitudes.

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Application +4 | -1

Two sound waves with wavelength 5m and 6m respectively, each propagate in the gas with velocity 330 m/s. We except the following number of beats per second

View answer and explanation

Correct option: D

11

Option D (11) is correct because the beat frequency is the absolute difference between the frequencies of the two waves.

Solution approach

Correct Answer + Core Reason: 11. The beat frequency is the difference between the two frequencies.

Step-by-Step Solution:

Step 1: Find the frequency of the first wave using the formula: $f_1 = \frac{v}{\lambda_1}$

Step 2: Find the frequency of the second wave using the formula: $f_2 = \frac{v}{\lambda_2}$

Step 3: Calculate the beat frequency using the formula: $f_{beat} = |f_1 - f_2|$

Step 4: Substitute the given values: $f_1 = \frac{330}{5} = 66 \text{ Hz}$ and $f_2 = \frac{330}{6} = 55 \text{ Hz}$

Step 5: Calculate the beat frequency: $f_{beat} = |66 - 55| = 11 \text{ Hz}$

Step 6: Final Answer: 11

Why other options are wrong:
- Option A: 6 is incorrect; it might be a miscalculation of the frequencies or their difference.
- Option B: 4 is incorrect; it might be a miscalculation of the frequencies or their difference.
- Option C: 9 is incorrect; it might be a miscalculation of the frequencies or their difference.

Key Concepts:
- Wave frequency: $f = \frac{v}{\lambda}$
- Beat frequency: $f_{beat} = |f_1 - f_2|$

Memory Aids:
- Beat frequency is always the absolute difference.

Quick Tips:
- Remember to calculate the individual frequencies before finding the beat frequency.

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Application +4 | -1

A transverse wave of amplitude

0.5\,\,m

and wavelength

1m

and frequency

2\,\,Hz

is propagating in a string in the negative

x

-direction. The expression for this wave is

View answer and explanation

Correct option: B

y\left( {x,t} \right) = 0.5\,\,\cos \left( {2\pi x + 4\pi t} \right)

Option B (

y(x,t) = 0.5\,\cos (2\pi x + 4\pi t)

) is correct because it represents a wave with the given amplitude, wavelength, and frequency propagating in the negative x-direction.

Solution approach

Correct Answer + Core Reason: $y(x,t) = 0.5\,\cos (2\pi x + 4\pi t)$ . This equation represents a wave with the given amplitude, wavelength, and frequency propagating in the negative x-direction.

Step-by-Step Solution:

Step 1: Identify the given information: Amplitude $A = 0.5\mathrm{ }\,m$ , wavelength $\lambda = 1\mathrm{ }\,m$ , frequency $f = 2\mathrm{ }\,Hz$ , and the wave is propagating in the negative x-direction.

Step 2: Recall the general form of a wave equation: $y(x,t) = A\cos(kx \pm \omega t)$ , where $k = \frac{2\pi}{\lambda}$ and $\omega = 2\pi f$ . The plus sign indicates propagation in the negative x-direction.

Step 3: Calculate the wave number: $k = \frac{2\pi}{\lambda} = \frac{2\pi}{1} = 2\pi$ .

Step 4: Calculate the angular frequency: $\omega = 2\pi f = 2\pi (2) = 4\pi$ .

Step 5: Substitute the values into the wave equation: $y(x,t) = 0.5\cos(2\pi x + 4\pi t)$ .

Step 6: Final Answer: The expression for the wave is $y(x,t) = 0.5\cos(2\pi x + 4\pi t)$ .

Why other options are wrong:
- **Option A ( $y(x,t) = 0.5\,\sin (2\pi x - 4\pi t)$ ): This is incorrect because it represents a wave propagating in the positive x-direction and uses a sine function instead of a cosine function (assuming the initial condition is a crest at x=0 and t=0).
- Option C ( $y(x,t) = 0.5\,\sin (\pi x - 2\pi t)$ ): This is incorrect because it has incorrect values for the wave number and angular frequency, and it propagates in the positive x-direction.
- Option D ( $y(x,t) = 0.5\,\cos (2\pi x + 2\pi t)$ ): This is incorrect because it has an incorrect value for the angular frequency.

Key Concepts:
- Transverse Wave Equation
- Amplitude
- Wavelength
- Frequency
- Wave Number
- Angular Frequency
- Direction of Propagation

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

Quick Tips:
- Always identify the amplitude, wavelength, and frequency correctly from the given information.
- Pay attention to the sign to determine the direction of propagation.
- Use the appropriate trigonometric function (sine or cosine) based on the initial conditions.

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Remembering +4 | -1

Water waves are

View answer and explanation

Correct option: C

Both longitudinal and transverse

Option C (Both longitudinal and transverse) is correct because water waves exhibit both longitudinal and transverse motion of water particles.

Solution approach

Correct Answer + Core Reason: Both longitudinal and transverse. Water waves exhibit both types of motion.

Step-by-Step Solution:

Step 1: In water waves, the water particles move in a circular or elliptical path.

Step 2: This motion has both a component parallel to the direction of wave propagation (longitudinal) and a component perpendicular to the direction of wave propagation (transverse).

Step 3: Therefore, water waves are a combination of both longitudinal and transverse waves.

Why other options are wrong:
- Option A: Longitudinal. This is incorrect because water waves also have a transverse component.
- Option B: Transverse. This is incorrect because water waves also have a longitudinal component.
- Option D: Neither longitudinal nor transverse. This is incorrect because water waves have both longitudinal and transverse components.

Key Concepts:
- Longitudinal Waves
- Transverse Waves
- Water Waves

Memory Aids:
- Remember that water waves are a combination of both longitudinal and transverse waves.

Quick Tips:
- Visualize the motion of water particles in a water wave to understand the combination of longitudinal and transverse motion.

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Remembering +4 | -1

The minimum audible wavelength at room temperature is about

View answer and explanation

Correct option: D

20\,\,mm

Option D (

20 \, mm

) is correct because the minimum audible frequency is around 20 Hz, and using the speed of sound (approximately 340 m/s), the corresponding wavelength is about 17 meters. The maximum audible frequency is around 20 kHz, which gives a wavelength of approximately 17 mm or roughly 20 mm.

Solution approach

Correct Answer + Core Reason: $20\\,\,mm$ . The minimum audible wavelength corresponds to the highest audible frequency and the speed of sound.

Step-by-Step Solution:

Step 1: Recall the relationship between wavelength, frequency, and speed of sound: $v = f\lambda$ .

Step 2: The audible frequency range is approximately 20 Hz to 20 kHz. We are looking for the *minimum* audible wavelength, which corresponds to the *maximum* audible frequency (20 kHz).

Step 3: Assume the speed of sound at room temperature is approximately 340 m/s.

Step 4: Calculate the minimum wavelength: $\lambda_{min} = \frac{v}{f_{max}} = \frac{340 \text{{ m/s}}}{20000 \text{{ Hz}}} = 0.017 \text{{ m}} = 17 \text{{ mm}}$ . This is closest to 20 mm.

Why other options are wrong:
- Option A: $0.2\\,{A^0}$ This is incorrect because Angstroms are units of length on the atomic scale, far smaller than audible wavelengths.
- Option B: $5\\,{A^0}$ This is incorrect because Angstroms are units of length on the atomic scale, far smaller than audible wavelengths.
- Option C: $\;5\\,\,cm\\,\,to\\,\,2\\,\,metre$ This is incorrect because while some audible wavelengths fall in this range, we are looking for the *minimum* audible wavelength.

Key Concepts:
- Relationship between speed, frequency, and wavelength: $v = f\lambda$
- Audible frequency range

Memory Aids:
- Remember the approximate speed of sound in air (340 m/s) and the audible frequency range (20 Hz - 20 kHz).

Quick Tips:
- Pay attention to whether the question asks for minimum or maximum wavelength/frequency.

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Understanding +4 | -1

Two waves having sinusoidal waveforms have different wavelengths and different amplitude. They will be having

View answer and explanation

Correct option: C

Different quality and different intensity

Option C is correct because different wavelengths and amplitudes result in different qualities and intensities of sound.

Solution approach

Correct Answer + Core Reason: Different quality and different intensity. Wavelength affects quality (timbre), and amplitude affects intensity.

Step-by-Step Solution:

Step 1: Define Pitch: Pitch is determined by frequency (or wavelength).

Step 2: Define Intensity: Intensity is determined by amplitude.

Step 3: Define Quality: Quality (or timbre) is determined by the combination of frequencies present in the sound wave.

Step 4: Analysis: Different wavelengths mean different frequencies, thus different pitch. Different amplitudes mean different intensities. Different waveforms (due to different combinations of frequencies) mean different qualities.

Why other options are wrong:
- Option A: Same pitch is incorrect because the wavelengths are different.
- Option B: Same quality is incorrect because the wavelengths and amplitudes are different.
- Option D: Same quality and different pitch are both incorrect because the wavelengths and amplitudes are different.

Key Concepts:
- Pitch
- Intensity
- Quality (Timbre)
- Wavelength
- Amplitude
- Frequency

Memory Aids:
- Pitch = Frequency
- Intensity = Amplitude
- Quality = Waveform

Quick Tips:
- Understand the relationship between physical properties of waves and perceived sound characteristics.

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Application +4 | -1

The disc of a siren containing

60

holes rotates at a constant speed of

360

rpm. The emitted sound is in unison with a tuning fork of frequency

View answer and explanation

Correct option: B

360 Hz

Option B (360 Hz) is correct because the frequency of the emitted sound matches the tuning fork frequency when calculated from the siren's rotation speed and number of holes.

Solution approach

Correct Answer + Core Reason: 360 Hz. The frequency is calculated from the rotation speed and number of holes.

Step-by-Step Solution:

Step 1: The siren has 60 holes, and it rotates at 360 rpm (revolutions per minute).

Step 2: Convert the rotation speed to revolutions per second:
$360 \text{{ rpm}} = \frac{{360}}{{60}} \text{{ revolutions per second}} = 6 \text{{ revolutions per second}}$

Step 3: Each revolution produces 60 sound pulses (one for each hole). Therefore, the frequency of the emitted sound is:
$f = 6 \text{{ revolutions per second}} \times 60 \text{{ holes per revolution}} = 360 \text{{ Hz}}$

Why other options are wrong:
- Option A: Incorrect due to incorrect calculation of the frequency.
- Option C: Incorrect due to incorrect calculation of the frequency.
- Option D: Incorrect due to incorrect calculation of the frequency.

Key Concepts:
- Frequency and rotation speed
- Relationship between holes and frequency

Memory Aids:
- Remember to convert rpm to revolutions per second.

Quick Tips:
- Multiply the number of holes by the rotation speed in revolutions per second to get the frequency.

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Understanding +4 | -1

The equation of a cylindrical progressive wave is

View answer and explanation

Correct option: C

y = \frac{a}{{\sqrt r }}\sin \left( {\omega t - kr} \right)

Option C is correct because the amplitude of a cylindrical wave decreases with the square root of the distance.

Solution approach

Correct Answer + Core Reason: $y = \frac{a}{{\sqrt r }}\sin \left( {\omega t - kr} \right)$ . The amplitude decreases with the square root of the distance in cylindrical waves.

Step-by-Step Solution:

Step 1: Understand Cylindrical Waves: Cylindrical waves propagate outwards from a line source.

Step 2: Amplitude Variation: The amplitude of a cylindrical wave decreases as the wave spreads out. The energy is distributed over a larger circumference, which increases linearly with radius $r$ . Since intensity is proportional to the square of amplitude, and intensity is inversely proportional to $r$ , the amplitude is inversely proportional to $\sqrt{r}$ .

Step 3: General Equation: The equation of a cylindrical wave includes a term $\frac{1}{\sqrt{r}}$ to account for the decreasing amplitude.

Why other options are wrong:
- Option A: $y = a\,\,\sin \,\,\omega t$ represents a simple harmonic motion, not a propagating wave.
- Option B: $y = a\,\,\sin \left( {\omega t - kr} \right)$ represents a plane wave with constant amplitude.
- Option D: $y = \frac{a}{r}\sin \left( {\omega t - kr} \right)$ represents a spherical wave, where amplitude decreases linearly with distance.

Key Concepts:
- Cylindrical Waves
- Amplitude
- Wave Propagation

Memory Aids:
- Cylindrical = Square Root (in amplitude)

Quick Tips:
- Remember the amplitude dependence on distance for different types of waves.

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Analyzing +4 | -1

The amplitude of a wave is given by

A

= \frac{c}{{a + b + c}}

. Resonance will occur when

View answer and explanation

Correct option: C

b = 0,a = c

Option C (

b = 0,a = c

) is correct because resonance occurs when the amplitude is maximum, which happens when the denominator is minimized.

Solution approach

Correct Answer + Core Reason: $b = 0,a = c$ . Resonance occurs when the amplitude is maximized, which requires minimizing the denominator.

Step-by-Step Solution:

Step 1: The amplitude is given by $A = \frac{c}{a + b + c}$ .

Step 2: Resonance occurs when the amplitude is maximum.

Step 3: To maximize the amplitude, the denominator $a + b + c$ must be minimized.

Step 4: If $b = 0$ and $a = c$ , then the denominator becomes $c + 0 + c = 2c$ .

Step 5: If $b = -\frac{c}{2}$ , then the denominator becomes $a - \frac{c}{2} + c = a + \frac{c}{2}$ . If $a$ is very small, this could be smaller than $2c$ , but in general, we want $a=c$ to minimize the denominator further.

Step 6: If $b = -\frac{a}{2}$ , then the denominator becomes $a - \frac{a}{2} + c = \frac{a}{2} + c$ . If $a$ is very small, this could be smaller than $2c$ , but in general, we want $a=c$ to minimize the denominator further.

Step 7: When $b = 0$ and $a = c$ , $A = \frac{c}{c + 0 + c} = \frac{c}{2c} = \frac{1}{2}$ . However, if $a$ approaches 0 and $b$ approaches $-c$ , the amplitude can become very large, approaching infinity. However, the question is asking for a condition where resonance *will* occur. The condition $b=0$ and $a=c$ is a valid condition for resonance, although not necessarily the only one.

Step 8: Final Answer: $b = 0,a = c$

Why other options are wrong:
- Option A: $b = - \frac{c}{2}$ is incorrect; it doesn't necessarily minimize the denominator for all values of $a$ .
- Option B: $b = - \frac{a}{2}$ is incorrect; it doesn't necessarily minimize the denominator for all values of $c$ .
- Option D: None of these is incorrect; option C is a valid condition for resonance.

Key Concepts:
- Resonance
- Amplitude

Memory Aids:
- Resonance occurs when amplitude is maximized.

Quick Tips:
- To maximize a fraction, minimize the denominator.

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Analyzing +4 | -1

The first overtone in a closed pipe has a frequency

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because the first overtone in a closed pipe is three times its fundamental frequency, which is not the same as or twice the fundamental frequency of an open tube of the same length.

Solution approach

Correct Answer + Core Reason: None of the above. The first overtone in a closed pipe is three times its fundamental frequency, which doesn't match the other options.

Step-by-Step Solution:

Step 1: Consider a closed pipe of length $L$ . The fundamental frequency is $f_1 = \frac{v}{4L}$ .

Step 2: The first overtone in a closed pipe is the next allowed frequency, which is $f_3 = \frac{3v}{4L} = 3f_1$ .

Step 3: Consider an open pipe of the same length $L$ . The fundamental frequency is $f'_1 = \frac{v}{2L}$ .

Step 4: Compare the first overtone of the closed pipe ( $f_3 = \frac{3v}{4L}$ ) with the fundamental frequency of the open pipe ( $f'_1 = \frac{v}{2L}$ ) and twice the fundamental frequency of the open pipe ( $2f'_1 = \frac{v}{L}$ ). None of the options match.

Why other options are wrong:
- Option A: Same as the fundamental frequency of an open tube of same length is incorrect. The first overtone of a closed pipe is three times its fundamental, while the fundamental of an open pipe is twice the fundamental of the closed pipe.
- Option B: Twice the fundamental frequency of an open tube of same length is incorrect. The first overtone of a closed pipe is $\frac{3}{2}$ times the fundamental of an open pipe of the same length.
- Option C: Same as that of the first overtone of an open tube of same length is incorrect. The first overtone of an open pipe is twice its fundamental, which is different from the first overtone of a closed pipe.

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 wave is represented by

y = 2\sin \pi \left( {4t - x} \right)

. where x and y are in metres and t is in seconds. The wavelength of the wave is

View answer and explanation

Correct option: C

2m

Option C (2m) is correct because the wavelength can be determined by comparing the given equation to the standard wave equation and extracting the coefficient of x.

Solution approach

Correct Answer + Core Reason: 2m. The wavelength is found by comparing the given wave equation to the standard form.

Step-by-Step Solution:

Step 1: Identify the given wave equation: $y = 2\sin \pi (4t - x)$ .

Step 2: Rewrite the equation as: $y = 2\sin (4\pi t - \pi x)$ .

Step 3: Compare this with the standard wave equation: $y = A\sin (\omega t - kx)$ , where $k = \frac{{2\pi}}{{\lambda}}$ .

Step 4: From the given equation, $k = \pi$ .

Step 5: Therefore, $\pi = \frac{{2\pi}}{{\lambda}}$ .

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

Why other options are wrong:
- Option A (1m): This is incorrect because it would imply $k = 2\pi$ , which doesn't match the given equation.
- **Option B ( $\frac{1}{2}m$ ):** This is incorrect because it would imply $k = 4\pi$ , which doesn't match the given equation.
- **Option D ( $\frac{3}{2}m$ ):** This is incorrect because it would imply $k = \frac{{4\pi}}{3}$ , which doesn't match the given equation.

Key Concepts:
- Wave equation: $y = A\sin (\omega t - kx)$
- Wave number: $k = \frac{{2\pi}}{{\lambda}}$

Memory Aids:
- Remember the standard form of a wave equation and the relationship between wave number and wavelength.

Quick Tips:
- Always compare the given equation with the standard form to identify the parameters.

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Understanding +4 | -1

If the tension and diameter of a sonometer wire having fun

View answer and explanation

Correct option: D

f^{\prime}=f

Option D (

f^{\prime}=f

) is correct because the frequency of a sonometer wire is independent of the diameter if the tension is kept constant and the length is unchanged.

Solution approach

Correct Answer + Core Reason: $f^{\prime}=f$ . The frequency is independent of the diameter if tension and length are constant.

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 $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.

Step 2: The linear mass density is given by $\mu = \frac{m}{L} = \frac{\rho V}{L} = \frac{\rho A L}{L} = \rho A = \rho \pi r^2 = \rho \pi (d/2)^2$ , where $\rho$ is the density, $A$ is the cross-sectional area, $r$ is the radius, and $d$ is the diameter.

Step 3: Substitute $\mu$ into the frequency formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\rho \pi (d/2)^2}} = \frac{1}{2Ld} \sqrt{\frac{4T}{\rho \pi}}$

Step 4: If the tension is made 9 times ( $T' = 9T$ ) and the diameter is made 3 times ( $d' = 3d$ ), the new frequency $f'$ is: $f' = \frac{1}{2L(3d)} \sqrt{\frac{4(9T)}{\rho \pi}} = \frac{1}{6Ld} \sqrt{\frac{36T}{\rho \pi}} = \frac{6}{6} \frac{1}{2Ld} \sqrt{\frac{4T}{\rho \pi}} = \frac{1}{2Ld} \sqrt{\frac{4T}{\rho \pi}} = f$

Step 5: Therefore, $f' = f$ .

Why other options are wrong:
- Option A: $f^{\prime}=3 f$ . This would be true if only the tension was changed, but the diameter also changes.
- Option B: $f^{\prime}=\frac{f}{3}$ . This is incorrect as the changes in tension and diameter cancel each other out.
- Option C: $f^{\prime}=9 f$ . This would be true if only the tension was changed, but the diameter also changes.

Key Concepts:
- Fundamental frequency of a string
- Linear mass density

Memory Aids:
- $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

Quick Tips:
- Remember to consider how both tension and diameter affect the frequency.

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Application +4 | -1

The amplitude of two waves are in ratio

5:2

. If all other conditions for the two waves Are same, then what is the ratio of their energy densities?

View answer and explanation

Correct option: D

25:4

Option D (

25:4

) is correct because the energy density is proportional to the square of the amplitude.

Solution approach

Correct Answer + Core Reason: $25:4$ . The ratio of energy densities is proportional to the square of the ratio of amplitudes.

Step-by-Step Solution:

Step 1: Identify the given information: The ratio of amplitudes is $A_1:A_2 = 5:2$ .

Step 2: Recall that the energy density of a wave is proportional to the square of its amplitude: $E \propto A^2$ .

Step 3: Calculate the ratio of energy densities: $\frac{E_1}{E_2} = \frac{A_1^2}{A_2^2} = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$ .

Why other options are wrong:
- Option A: $5:2$ is incorrect. It is the ratio of the amplitudes, not the energy densities.
- Option B: $5:4$ is incorrect. It might be a result of taking the square root of the ratio of amplitudes or using an incorrect formula.
- Option C: $4:5$ is incorrect. It is the inverse of a possible incorrect calculation.

Key Concepts:
- Energy density of a wave
- Amplitude of a wave
- Proportionality between energy density and amplitude squared

Memory Aids:
- Remember that energy density is proportional to the square of the amplitude.

Quick Tips:
- Always pay attention to the relationship between energy density and amplitude.

119 Medium

119 of 159Question permalink

Analyzing +4 | -1

Assertion (A) : When a wave travels from a denser medium to rarer medium, then amplitude of oscillation increases.

Reason (R) : In denser medium, speed of wave is less compared to a rarer medium.

View answer and explanation

Correct option: B

If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the speed of the wave doesn't directly explain the amplitude increase.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. The amplitude increases due to energy conservation, and the speed of the wave is indeed less in a denser medium, but the speed doesn't directly cause the amplitude increase.

Step-by-Step Solution:

Step 1: Analyze the Assertion:\nWhen a wave travels from a denser medium to a rarer medium, the amplitude of oscillation increases. This is generally true because to conserve energy, if the wave spreads out more in the rarer medium, the amplitude must increase.

Step 2: Analyze the Reason:\nIn a denser medium, the speed of the wave is less compared to a rarer medium. This is also true because the speed of a wave depends on the properties of the medium, and denser mediums typically have lower wave speeds.

Step 3: Determine if the Reason is the correct explanation for the Assertion:\nThe reason why the amplitude increases is due to the conservation of energy. When a wave travels from a denser to a rarer medium, its energy spreads over a larger area, causing the amplitude to increase. The change in speed is a separate phenomenon and not the direct cause of the amplitude increase.

Why other options are wrong:
- Option A: If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. The reason is true, but it is not the correct explanation for the assertion.
- Option C: If Assertion is true, but the Reason is false. Both assertion and reason are true.
- Option D: If Assertion is false but the Reason is true. The assertion is true.

Key Concepts:
- Wave propagation
- Amplitude of a wave
- Speed of a wave in different mediums
- Conservation of energy

Memory Aids:
- Remember that the speed of a wave depends on the medium, and the amplitude is related to the energy of the wave.

Quick Tips:
- When analyzing assertion-reason questions, first determine if both statements are true. Then, determine if the reason directly explains the assertion.

120 Medium

120 of 159Question permalink

Application +4 | -1

An observer moves towards a stationary source of sound with a velocity one-fourth of the velocity of sound. What is the percentage increases in the apparent frequency?

View answer and explanation

Correct option: A

25\%

Option A (

25\%

) is correct because the apparent frequency increases by 25% when the observer moves towards the stationary source at one-fourth the speed of sound.

Solution approach

Correct Answer + Core Reason: $25\%$ . The apparent frequency increases due to the Doppler effect.

Step-by-Step Solution:

Step 1: Let $v$ be the velocity of sound. The observer moves towards the stationary source with a velocity $v_o = \frac{v}{4}$ .

Step 2: The apparent frequency $f'$ is given by the Doppler effect formula: $f' = f \times \frac{{v + v_o}}{{v}}$ , where $f$ is the original frequency.

Step 3: Substitute $v_o = \frac{v}{4}$ into the formula:
$f' = f \times \frac{{v + \frac{v}{4}}}{{v}} = f \times \frac{{\frac{5v}{4}}}{{v}} = f \times \frac{5}{4} = 1.25f$

Step 4: The percentage increase in the apparent frequency is:
$\frac{{f' - f}}{{f}} \times 100 = \frac{{1.25f - f}}{{f}} \times 100 = \frac{{0.25f}}{{f}} \times 100 = 0.25 \times 100 = 25\%$

Why other options are wrong:
- Option B: Incorrect due to incorrect calculation of the percentage increase.
- Option C: Incorrect due to incorrect calculation of the percentage increase.
- Option D: Incorrect due to incorrect calculation of the percentage increase.

Key Concepts:
- Doppler Effect
- Frequency shift due to relative motion

Memory Aids:
- Remember the Doppler effect formula.

Quick Tips:
- Pay attention to the direction of motion and use the correct sign in the Doppler effect formula.

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