WAVES MCQs for NEET UG PHYSICS — Medium, Page 6

151 Medium

Remembering +4 | -1

In the equation

y=A \sin (k x-\omega t+\phi_0),

the term phase is defined as

View answer and explanation

Correct option: D

k x-\omega t+\phi_0

Option D (

k x-\omega t+\phi_0

) is correct because the phase in a wave equation encompasses all terms inside the trigonometric function.

Solution approach

Correct Answer + Core Reason: $k x-\omega t+\phi_0$ . The phase includes all terms inside the sine function.

Step-by-Step Solution:

Step 1: Identify the general form of a wave equation: $y=A \sin (k x-\omega t+\phi_0)$ .

Step 2: The phase is the argument of the sine function, which is the entire expression inside the parentheses.

Step 3: Therefore, the phase is $k x-\omega t+\phi_0$ .

Why other options are wrong:
- Option A: $\phi_0$ is incorrect because it only represents the initial phase constant, not the entire phase.
- Option B: $\phi_0=\omega t$ is incorrect because it's not a general definition and doesn't include the spatial term $kx$ .
- Option C: $k x+\phi_0$ is incorrect because it doesn't include the time-dependent term $-\omega t$ .

Key Concepts:
- Wave equation
- Phase of a wave

Memory Aids:
- Phase is everything inside the trigonometric function.

Quick Tips:
- Remember that the phase is a function of both position (x) and time (t).

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Application +4 | -1

A string under a tension of

129.6 \mathrm{~N}

produces 10 beats/s when it is vibrated along with a tuning fork. When the tension in the string is increased to

160 \mathrm{~N}

, it sounds in unision with the same tuning fork. Calculate the fundamental frequency of the tuning fork.

View answer and explanation

Correct option: A

100

Option A (100) is correct because the fundamental frequency of the tuning fork is calculated by considering the change in frequency of the string due to the change in tension and the resulting beat frequency.

Solution approach

Correct Answer + Core Reason: 100 Hz. The frequency of the tuning fork is determined by analyzing the beats produced when the string's frequency is close to the tuning fork's frequency.

Step-by-Step Solution:

Step 1: Let $f_t$ be the frequency of the tuning fork. Let $f_1$ be the frequency of the string with tension $T_1 = 129.6 \, N$ , and $f_2$ be the frequency of the string with tension $T_2 = 160 \, N$ .

Step 2: We are given that the beat frequency with $T_1$ is 10 Hz. This means $|f_t - f_1| = 10$ . So, $f_t = f_1 \pm 10$ .

Step 3: When the tension is increased to $T_2$ , the string sounds in unison with the tuning fork, meaning $f_t = f_2$ .

Step 4: The frequency of a string is proportional to the square root of the tension: $f \propto \sqrt{{T}}$ . Therefore, $\frac{{f_1}}{{f_2}} = \sqrt{{\frac{{T_1}}{{T_2}}}} = \sqrt{{\frac{{129.6}}{{160}}}} = \sqrt{{\frac{{1296}}{{1600}}}} = \frac{{36}}{{40}} = \frac{{9}}{{10}}$ .

Step 5: So, $f_1 = \frac{{9}}{{10}} f_2 = \frac{{9}}{{10}} f_t$ .

Step 6: Substitute this into the beat frequency equation: $f_t = \frac{{9}}{{10}} f_t \pm 10$ .

Step 7: Consider the two cases:
* Case 1: $f_t = \frac{{9}}{{10}} f_t + 10$ . Then $\frac{{1}}{{10}} f_t = 10$ , so $f_t = 100 \, Hz$ .
* Case 2: $f_t = \frac{{9}}{{10}} f_t - 10$ . Then $\frac{{1}}{{10}} f_t = -10$ , so $f_t = -100 \, Hz$ . Since frequency cannot be negative, this case is invalid.

Step 8: Therefore, the frequency of the tuning fork is $f_t = 100 \, Hz$ .

Final Answer: The fundamental frequency of the tuning fork is 100 Hz.

Why other options are wrong:
- Option B: 0: This is incorrect as it would imply no sound from the tuning fork.
- Option C: 0: This is incorrect as it would imply no sound from the tuning fork.
- Option D: 0: This is incorrect as it would imply no sound from the tuning fork.

Key Concepts:
- Fundamental frequency of a string
- Relationship between frequency and tension
- Beat frequency

Memory Aids:
- $f \propto \sqrt{{T}}$ (frequency is proportional to the square root of tension).
- Beat frequency = $|f_1 - f_2|$

Quick Tips:
- Remember the formula for the fundamental frequency of a vibrating string.
- Use the beat frequency to relate the frequencies of the string and the tuning fork.
- Consider both possible cases when dealing with beat frequencies.

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Remembering +4 | -1

The harmonic which are present in a pipe open at one end are

View answer and explanation

Correct option: A

Odd harmonics

Option A (Odd harmonics) is correct because a pipe open at one end only supports odd multiples of the fundamental frequency due to the boundary conditions.

Solution approach

Correct Answer + Core Reason: Odd harmonics. A pipe open at one end only supports odd multiples of the fundamental frequency.

Step-by-Step Solution:

Step 1: Consider a pipe open at one end. The boundary conditions require a node at the closed end and an antinode at the open end.

Step 2: The fundamental frequency (first harmonic) corresponds to a quarter of a wavelength fitting into the pipe: $L = \frac{\lambda_1}{4}$ , so $\lambda_1 = 4L$ .

Step 3: The next possible harmonic (third harmonic) corresponds to three-quarters of a wavelength fitting into the pipe: $L = \frac{3\lambda_3}{4}$ , so $\lambda_3 = \frac{4L}{3}$ .

Step 4: The next possible harmonic (fifth harmonic) corresponds to five-quarters of a wavelength fitting into the pipe: $L = \frac{5\lambda_5}{4}$ , so $\lambda_5 = \frac{4L}{5}$ .

Step 5: In general, the wavelengths are given by $\lambda_n = \frac{4L}{n}$ , where $n$ is an odd integer (1, 3, 5, ...). Therefore, only odd harmonics are present.

Step 6: Final Answer: Odd harmonics

Why other options are wrong:
- Option B (Even harmonics): Even harmonics are not present in a pipe open at one end.
- Option C (Even as well as odd harmonics): Only odd harmonics are present.
- Option D (None of these): Odd harmonics are present.

Key Concepts:
- Harmonics
- Standing waves
- Pipes open at one end

Memory Aids:
- Remember that a pipe open at one end only supports odd harmonics.

Quick Tips:
- The boundary conditions determine which harmonics are present.

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Remembering +4 | -1

The waves associated with the motion of microscopic particles such as electrons, protons, neutrons are called

View answer and explanation

Correct option: A

Matter waves

Option A is correct because matter waves are associated with the motion of microscopic particles.

Solution approach

Correct Answer + Core Reason: Matter waves. These waves are associated with the motion of microscopic particles.

Step-by-Step Solution:

Step 1: Define Matter Waves: Matter waves, also known as de Broglie waves, are associated with particles having mass and momentum.

Step 2: Microscopic Particles: Electrons, protons, and neutrons are microscopic particles that exhibit wave-like behavior.

Step 3: Application: The wave nature of these particles is described by matter waves.

Why other options are wrong:
- Option B: Mechanical waves require a medium to propagate, while matter waves do not necessarily need a medium.
- Option C: Electromagnetic waves are composed of oscillating electric and magnetic fields and are not directly associated with the motion of microscopic particles in the same way as matter waves.
- Option D: Cosmic waves is not a standard term in physics.

Key Concepts:
- Matter Waves (de Broglie waves)
- Wave-particle duality

Memory Aids:
- Matter = Microscopic Particles

Quick Tips:
- Remember de Broglie's hypothesis: All matter exhibits wave-like properties.

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Understanding +4 | -1

Out of the given waves (1), (2), (3) and (4)

y = a\sin \left( {kx + \omega t} \right)

...(

1

)

y = a\sin \left( {\omega t - kx} \right)

…(

2

)

y = a\cos \left( {kx + \omega t} \right)

…(

3

)

y = a\cos \left( {\omega t - kx} \right)

…(

4

)
Emitted by four different sources

{S_1},{S_2},{S_3}\,\,and\,\,{S_4}

respectively, interference phenomena would be observed in space under appropriate conditions when

View answer and explanation

Correct option: C

Source

{S_2}

emits wave (

2

) and 

{S_4}

 emits wave (

4

)

Option C (Source

{S_2}

emits wave (

2

) and 

{S_4}

 emits wave (

4

)) is correct because for interference to occur, the waves must be coherent, meaning they have a constant phase difference. Waves (2) and (4) are both traveling in the +x direction and have the same form, thus they can interfere.

Solution approach

Correct Answer + Core Reason: Source ${S_2}$ emits wave ( $2$ ) and  ${S_4}$  emits wave ( $4$ ). For interference, waves should be coherent and have a constant phase relationship.

Step-by-Step Solution:

Step 1: Recall the conditions for interference.

Step 2: For interference to occur, the waves must be coherent (same frequency and constant phase difference) and traveling in the same direction.

Step 3: Analyze the given wave equations:
- Wave (1): $y = a\sin(kx + \omega t)$ - Traveling in the -x direction
- Wave (2): $y = a\sin(\omega t - kx)$ - Traveling in the +x direction
- Wave (3): $y = a\cos(kx + \omega t)$ - Traveling in the -x direction
- Wave (4): $y = a\cos(\omega t - kx)$ - Traveling in the +x direction

Step 4: Compare the waves in each option:
- Option A: Waves (1) and (2) are traveling in opposite directions.
- Option B: Waves (3) and (4) are traveling in opposite directions.
- Option C: Waves (2) and (4) are traveling in the same direction and have a constant phase relationship (both are sinusoidal).
- Option D: Waves (4) and (3) are traveling in opposite directions.

Step 5: Only option C has waves traveling in the same direction with a constant phase relationship.

Why other options are wrong:
- Option A: Sources ${S_1}$ emits wave ( $1$ ) and ${S_2}$ emits wave ( $2$ ). These waves travel in opposite directions, so interference is less likely.
- Option B: Source ${S_3}$ emits wave ( $3$ ) and ${S_4}$ emits wave ( $4$ ). These waves travel in opposite directions, so interference is less likely.
- Option D: ${S_4}$ emits waves ( $4$ ) and ${S_3}$ emits waves ( $3$ ). These waves travel in opposite directions, so interference is less likely.

Key Concepts:
- Interference of waves
- Coherence
- Wave direction

Memory Aids:
- Remember that coherent waves are needed for sustained interference.

Quick Tips:
- Pay attention to the sign in front of $kx$ to determine the direction of wave propagation.

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Understanding +4 | -1

The musical interval between two tones of frequencies

320\,\,Hz

and

240\,\,Hz

is

View answer and explanation

Correct option: B

\left( {\frac{4}{3}} \right)

Option B ((4/3)) is correct because the musical interval is the ratio of the two frequencies.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{4}{3}} \right)$ . The musical interval is the ratio of the two frequencies.

Step-by-Step Solution:

Step 1: Define musical interval.\nThe musical interval between two tones is the ratio of their frequencies.

Step 2: Calculate the ratio.\nMusical interval = $\frac{320}{240} = \frac{4}{3}$

Why other options are wrong:
- Option A: 80 is the difference between the frequencies, not the ratio.
- Option C: 560 is the sum of the frequencies, not the ratio.
- Option D: $320 \times 240$ is the product of the frequencies, not the ratio.

Key Concepts:
- Musical interval
- Frequency ratio

Memory Aids:
- Musical interval is a ratio.

Quick Tips:
- Divide the higher frequency by the lower frequency to find the musical interval.

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Application +4 | -1

A wave travelling along the x-axis is described by the equation

y\left( {x,t} \right) = 0.005\,\,\,cos\left( {\alpha x - \beta t} \right)

. If the wavelength and the time period of the wave are

0.08{\mkern 1mu} {\mkern 1mu} \,\,m\,\,\,and\,\,2.0{\mkern 1mu} {\mkern 1mu} s

, respectively, than

\alpha \,\,and\,\beta

in appropriate unit are

View answer and explanation

Correct option: A

a = 25.00\pi ,\mathrm{ }\beta \mathrm{ }\pi

Option A (

a = 25.00\pi ,\mathrm{ }\beta \mathrm{ }\pi

) is correct because it correctly calculates the values of

\alpha

and

\beta

using the given wavelength and time period.

Solution approach

Correct Answer + Core Reason: $a = 25.00\pi ,\mathrm{ }\beta \mathrm{ }\pi$ . The values of $\alpha$ and $\beta$ are calculated using the given wavelength and time period.

Step-by-Step Solution:

Step 1: Identify the given information: The equation of the wave is $y(x,t) = 0.005\,\cos(\alpha x - \beta t)$ , the wavelength is $\lambda = 0.08\mathrm{ }\,m$ , and the time period is $T = 2.0\mathrm{ }\,s$ .

Step 2: Recall the relationship between wave number ( $k$ ) and wavelength: $k = \frac{2\pi}{\lambda}$ .

Step 3: Recall the relationship between angular frequency ( $\omega$ ) and time period: $\omega = \frac{2\pi}{T}$ .

Step 4: From the given equation, identify that $\alpha = k$ and $\beta = \omega$ .

Step 5: Calculate $\alpha$ : $\alpha = \frac{2\pi}{\lambda} = \frac{2\pi}{0.08} = 25\pi$ .

Step 6: Calculate $\beta$ : $\beta = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi$ .

Step 7: Final Answer: $\alpha = 25\pi$ and $\beta = \pi$ .

Why other options are wrong:
- **Option B ( $\alpha = \frac{{0.08}}{\pi },\beta = \frac{{2.0}}{\pi }$ ):** This is incorrect because it incorrectly calculates $\alpha$ and $\beta$ by inverting the relationships with wavelength and time period.
- **Option C ( $\alpha = \frac{{0.04}}{\pi },\beta = \frac{{1.0}}{\pi }$ ):** This is incorrect because it incorrectly calculates $\alpha$ and $\beta$ by using half the wavelength and half the time period.
- **Option D ( $\alpha = 12.5\pi ,\beta = \frac{\pi }{{2.0}}$ ):** This is incorrect because it incorrectly calculates $\alpha$ and $\beta$ by using half the correct value for $\alpha$ and half the correct value for $\beta$ .

Key Concepts:
- Wave Equation
- Wavelength
- Time Period
- Wave Number
- Angular Frequency

Memory Aids:
- Remember the relationships: $k = \frac{2\pi}{\lambda}$ and $\omega = \frac{2\pi}{T}$

Quick Tips:
- Always identify the wavelength and time period correctly from the given information.
- Ensure the correct formulas are used relating wave number, angular frequency, wavelength, and time period.

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Application +4 | -1

A sonometer wire has a length

114 \mathrm{~cm}

between two fixed ends. Where should two bridges be placed, so as to divide the wire into three segments whose fundamental frequencies are in the ratio

1: 3: 4

?

View answer and explanation

Correct option: A

l_1=72 \mathrm{~cm}, l_2=24 \mathrm{~cm}, l_3=18 \mathrm{~cm}

Option A (

l_1=72 \mathrm{~cm}, l_2=24 \mathrm{~cm}, l_3=18 \mathrm{~cm}

) is correct because the lengths satisfy both the total length condition and the inverse frequency ratio condition.

Solution approach

Correct Answer + Core Reason: $l_1=72 \mathrm{~cm}, l_2=24 \mathrm{~cm}, l_3=18 \mathrm{~cm}$ . The lengths are calculated based on the inverse relationship between frequency and length for a vibrating string, and the total length of the wire.

Step-by-Step Solution:

Step 1: The total length of the wire is 114 cm: $l_1 + l_2 + l_3 = 114$ .

Step 2: The fundamental frequencies are in the ratio 1:3:4. Since frequency is inversely proportional to length, the lengths are in the ratio $\frac{{1}}{{1}} : \frac{{1}}{{3}} : \frac{{1}}{{4}}$ .

Step 3: Let $l_1 = k$ , $l_2 = \frac{{k}}{{3}}$ , and $l_3 = \frac{{k}}{{4}}$ .

Step 4: Substitute these into the total length equation: $k + \frac{{k}}{{3}} + \frac{{k}}{{4}} = 114$ .

Step 5: Find a common denominator and solve for $k$ : $\frac{{12k + 4k + 3k}}{{12}} = 114$ , which simplifies to $\frac{{19k}}{{12}} = 114$ .

Step 6: Solve for $k$ : $k = \frac{{114 \times 12}}{{19}} = 6 \times 12 = 72$ .

Step 7: Calculate the lengths of the segments: $l_1 = 72 \, cm$ , $l_2 = \frac{{72}}{{3}} = 24 \, cm$ , and $l_3 = \frac{{72}}{{4}} = 18 \, cm$ .

Final Answer: The lengths of the three segments are $l_1 = 72 \, cm$ , $l_2 = 24 \, cm$ , and $l_3 = 18 \, cm$ .

Why other options are wrong:
- Option B: $l_1=60 \mathrm{~cm}, l_2=40 \mathrm{~cm}, l_3=14 \mathrm{~cm}$ : The sum of these lengths is 114 cm, but they do not satisfy the frequency ratio.
- Option C: $l_1=52 \mathrm{~cm}, l_2=30 \mathrm{~cm}, l_3=32 \mathrm{~cm}$ : The sum of these lengths is 114 cm, but they do not satisfy the frequency ratio.
- Option D: $l_1=65 \mathrm{~cm}, l_2=30 \mathrm{~cm}, l_3=19 \mathrm{~cm}$ : The sum of these lengths is 114 cm, but they do not satisfy the frequency ratio.

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

Which of the following is not true for the progressive wave

y=4 \sin 2 \pi\left(\frac{t}{0.02}-\frac{x}{100}\right)

where

x

and

y

are in

\mathrm{cm}

and

t

in second?

View answer and explanation

Correct option: D

The velocity of wave propagation is

2 \mathrm{~cm} / \mathrm{s}

Option D (Wave 2 has greatest wave speed and wave 3 has maximum transverse string speed) is correct because wave speed is

\omega/k

and transverse speed is

A\omega

.

Solution approach

Correct Answer + Core Reason: Wave 2 has greatest wave speed and wave 3 has maximum transverse string speed. Wave speed is $\omega/k$ and transverse speed is $A\omega$ .

Step-by-Step Solution:

Step 1: Identify the given information: The three wave equations are:
- $y_1 = (2\mathrm{ }\,cm)\sin(3x - 6t)$
- $y_2 = (3\mathrm{ }\,cm)\sin(4x - 12t)$
- $y_3 = (4\mathrm{ }\,cm)\sin(5x - 11t)$

Step 2: Calculate the wave speed for each wave: $v = \frac{\omega}{k}$ .
- $v_1 = \frac{6}{3} = 2\mathrm{ }\,cm/s$
- $v_2 = \frac{12}{4} = 3\mathrm{ }\,cm/s$
- $v_3 = \frac{11}{5} = 2.2\mathrm{ }\,cm/s$
- Therefore, wave 2 has the greatest wave speed.

Step 3: Calculate the maximum transverse string speed for each wave: $v_{max} = A\omega$ .
- $(v_{max})_1 = 2 \times 6 = 12\mathrm{ }\,cm/s$
- $(v_{max})_2 = 3 \times 12 = 36\mathrm{ }\,cm/s$
- $(v_{max})_3 = 4 \times 11 = 44\mathrm{ }\,cm/s$
- Therefore, wave 3 has the greatest maximum transverse string speed.

Step 4: Final Answer: Wave 2 has the greatest wave speed, and wave 3 has the greatest maximum transverse string speed.

Why other options are wrong:
- Option A (Wave 1 has greatest wave speed and wave has maximum transverse string speed):

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