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NEET PHYSICSWAVEEasy

Question

The wave described by y=0.25sin(10πx2πt)y = 0.25\sin(10\pi x - 2\pi t), where xx and yy are in metres and tt in seconds, is a wave traveling along the:

A

-ve xx direction with frequency 1 Hz1 \text{ Hz}.

B

+ve xx direction with frequency π Hz\pi \text{ Hz} and wavelength λ=0.2 m\lambda = 0.2 \text{ m}.

C

+ve xx direction with frequency 1 Hz1 \text{ Hz} and wavelength λ=0.2 m\lambda = 0.2 \text{ m}.

D

-ve xx direction with amplitude 0.25 m0.25 \text{ m} and wavelength λ=0.2 m\lambda = 0.2 \text{ m}.

Step-by-Step Solution

  1. Compare with standard wave equation: The given wave equation is y=0.25sin(10πx2πt)y = 0.25\sin(10\pi x - 2\pi t). The general equation for a progressive wave is y=Asin(kxωt)y = A\sin(kx - \omega t).
  2. Determine Direction of Propagation: The negative sign between the kxkx and ωt\omega t terms indicates that the wave is travelling in the positive xx-direction (+ve xx) .
  3. Calculate Frequency (ff): From the equation, angular frequency ω=2π\omega = 2\pi. Since ω=2πf\omega = 2\pi f, we have 2πf=2π2\pi f = 2\pi, which gives f=1 Hzf = 1 \text{ Hz} .
  4. Calculate Wavelength (λ\lambda): From the equation, wave number k=10πk = 10\pi. Since k=2πλk = \frac{2\pi}{\lambda}, we have 2πλ=10π\frac{2\pi}{\lambda} = 10\pi, which gives λ=210=0.2 m\lambda = \frac{2}{10} = 0.2 \text{ m} . Thus, the wave travels in the +ve xx direction with a frequency of 1 Hz1 \text{ Hz} and wavelength of 0.2 m0.2 \text{ m}.

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.

PHYSICSWAVEdescribedmetressecondstraveling

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