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Waves question

2006 · Q171
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Waves question

2006 · Q171

NEETPhysicsWavesMCQ+4 / −1
A transverse wave propagating along x-axis is represented by y(x, t) = 8.0 sin (0.5 π\piπx −-− 4π\piπt −-− π\piπ/4) where x is in metres and t is in seconds. The speed of the wave is
  1. A
    8 m/s
  2. B
    4π\piπ m/s
  3. C
    0.5π\piπ m/s
  4. D
    π\piπ/4 m/s.
View written solutionFree

Correct answer: A

y(x,t)=8.0sin⁡(0.5πx−4πt−π4)y\left( {x,t} \right) = 8.0\sin \left( {0.5\pi x - 4\pi t - {\pi \over 4}} \right)y(x,t)=8.0sin(0.5πx−4πt−4π​)

Compare with a standard wave equation,

y=asin⁡(2πxλ−2πtT+ϕ)y = a\sin \left( {{{2\pi x} \over \lambda } - {{2\pi t} \over T} + \phi } \right)y=asin(λ2πx​−T2πt​+ϕ)

we get 2πλ=0.5π⇒λ=2π0.5π=4m{{2\pi } \over \lambda } = 0.5\pi \Rightarrow \lambda = {{2\pi } \over {0.5\pi }} = 4mλ2π​=0.5π⇒λ=0.5π2π​=4m

2πT=4π⇒T=2π4π=12sec⁡{{2\pi } \over T} = 4\pi \Rightarrow T = {{2\pi } \over {4\pi }} = {1 \over 2}\sec T2π​=4π⇒T=4π2π​=21​sec

υ=1/T=2 Hz\upsilon = 1/T = 2\,Hzυ=1/T=2Hz.

Wave velocity, v=λυ=4×2=8v = \lambda \upsilon = 4 \times 2 = 8v=λυ=4×2=8 m/sec.

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