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NEET PHYSICSWAVE OPTICSMedium

Question

A wave in a string has an amplitude of 2 cm2 \text{ cm}. The wave travels in the positive direction of the x-axis with a speed of 128 m/s128 \text{ m/s} and it is noted that 55 complete waves fit in the 4 m4 \text{ m} length of the string. The equation describing the wave is:

A

y=(0.02 m)sin(7.85x+1005t)y=(0.02 \text{ m})\sin(7.85x+1005t)

B

y=(0.02 m)sin(15.7x2010t)y=(0.02 \text{ m})\sin(15.7x-2010t)

C

y=(0.02 m)sin(15.7x+2010t)y=(0.02 \text{ m})\sin(15.7x+2010t)

D

y=(0.02 m)sin(7.85x1005t)y=(0.02 \text{ m})\sin(7.85x-1005t)

Step-by-Step Solution

Given: Amplitude, A=2 cm=0.02 mA = 2 \text{ cm} = 0.02 \text{ m} Velocity of wave, v=128 m/sv = 128 \text{ m/s} 55 complete waves fit in a 4 m4 \text{ m} length. Therefore, the wavelength λ\lambda is: λ=45=0.8 m\lambda = \frac{4}{5} = 0.8 \text{ m} The wave number kk is calculated as: k=2πλ=2×3.140.8=7.85 m1k = \frac{2\pi}{\lambda} = \frac{2 \times 3.14}{0.8} = 7.85 \text{ m}^{-1} The angular frequency ω\omega is calculated using the relation v=ωkv = \frac{\omega}{k}: ω=v×k=128×7.85=1004.81005 rad/s\omega = v \times k = 128 \times 7.85 = 1004.8 \approx 1005 \text{ rad/s} Since the wave is traveling in the positive x-direction, the general equation of the wave is of the form: y=Asin(kxωt)y = A \sin(kx - \omega t) Substituting the derived values: y=(0.02 m)sin(7.85x1005t)y = (0.02 \text{ m}) \sin(7.85x - 1005t)

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. 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.

PHYSICSWAVE OPTICSstringamplitudetravelspositivedirection

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