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NEET PHYSICSWAVEEasy

Question

A wave travelling in the positive x-direction having maximum displacement along y-direction as 1 m1 \text{ m}, wavelength 2π m2\pi \text{ m} and frequency of 1/π Hz1/\pi \text{ Hz} is represented by

A

y=sin(x2t)y=\sin(x-2t)

B

y=sin(2πx2πt)y=\sin(2\pi x-2\pi t)

C

y=sin(10πx20πt)y=\sin(10\pi x-20\pi t)

D

y=sin(2πx+2πt)y=\sin(2\pi x+2\pi t)

Step-by-Step Solution

  1. Identify the given parameters: Maximum displacement (Amplitude), A=1 mA = 1 \text{ m}; Wavelength, λ=2π m\lambda = 2\pi \text{ m}; Frequency, f=1π Hzf = \frac{1}{\pi} \text{ Hz}.
  2. Calculate wave number (kk): The wave number is given by k=2πλk = \frac{2\pi}{\lambda}. k=2π2π=1 m1k = \frac{2\pi}{2\pi} = 1 \text{ m}^{-1}
  3. Calculate angular frequency (ω\omega): The angular frequency is given by ω=2πf\omega = 2\pi f. ω=2π×1π=2 rad/s\omega = 2\pi \times \frac{1}{\pi} = 2 \text{ rad/s}
  4. Formulate the wave equation: The general equation for a progressive wave travelling in the positive x-direction is y=Asin(kxωt)y = A \sin(kx - \omega t) or y=Asin(ωtkx)y = A \sin(\omega t - kx) . Substituting the calculated values into the first form: y=1sin(1x2t)=sin(x2t)y = 1 \cdot \sin(1 \cdot x - 2 \cdot t) = \sin(x - 2t)

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVEtravellingpositivexdirectionhavingmaximum

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If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

A.$1:2$
B.$1:1$
C.$\sqrt{2}:1$
D.$1:\sqrt{2}$
EasySolve

When a string is divided into three segments of lengths $l_1, l_2$ and $l_3$, the fundamental frequencies of these three segments are $\nu_1, \nu_2$ and $\nu_3$ respectively. The original fundamental frequency ($\nu$) of the string is:

A.$\sqrt{\nu}=\sqrt{\nu_1}+\sqrt{\nu_2}+\sqrt{\nu_3}$
B.$\nu=\nu_1+\nu_2+\nu_3$
C.$\frac{1}{\nu}=\frac{1}{\nu_1}+\frac{1}{\nu_2}+\frac{1}{\nu_3}$
D.$\frac{1}{\sqrt{\nu}}=\frac{1}{\sqrt{\nu_1}}+\frac{1}{\sqrt{\nu_2}}+\frac{1}{\sqrt{\nu_3}}$
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Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:

A.$I_1+I_2$
B.$(\sqrt{I_1}+\sqrt{I_2})^2$
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D.$2(I_1+I_2)$
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The two nearest harmonics of a tube closed at one end and open at the other end are $220 \text{ Hz}$ and $260 \text{ Hz}$. What is the fundamental frequency of the system?

A.$10 \text{ Hz}$
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C.$30 \text{ Hz}$
D.$40 \text{ Hz}$
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A transverse wave propagating along the $x$-axis is represented by: $y(x,t) = 8.0\sin(0.5\pi x - 4\pi t - \frac{\pi}{4})$, where $x$ is in meters and $t$ in seconds. The speed of the wave is:

A.$4\pi \text{ m/s}$
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D.$8 \text{ m/s}$
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The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is $20 \text{ cm}$, the length of the open organ pipe is:

A.$13.2 \text{ cm}$
B.$8 \text{ cm}$
C.$12.5 \text{ cm}$
D.$16 \text{ cm}$
MediumSolve

The number of possible natural oscillations of the air column in a pipe closed at one end of a length of $85 \text{ cm}$ whose frequencies lie below $1250 \text{ Hz}$ is: (velocity of sound $340 \text{ ms}^{-1}$)

A.4
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A source of unknown frequency gives $4 \text{ beats/s}$ when sounded with a source of known frequency $250 \text{ Hz}$. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 \text{ Hz}$. The unknown frequency is

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