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Waves question

2012 · Q96
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Waves question

2012 · Q96

NEETPhysicsWavesMCQ+4 / −1
The equation of a simple harmonic wave is given by

y = 3 sinπ2{\pi \over 2}2π​(50t −-− x),

where x and y are in metres and t is in seconds. The ratio of maximum particle velocity to the wave velocity is
  1. A
    2π\piπ
  2. B
    32π{3 \over 2}\pi23​π
  3. C
    3π3\pi3π
  4. D
    23π{2 \over 3}\pi32​π
View written solutionFree

Correct answer: B

y=3sin⁡π2(50t−x)y = 3\sin {\pi \over 2}\left( {50t - x} \right)y=3sin2π​(50t−x)

y=3sin⁡(25πt−π2x)y = 3\sin \left( {25\pi t - {\pi \over 2}x} \right)y=3sin(25πt−2π​x) on comparing with the standard wave equation

y=asin⁡(ωt−kx)y = a\sin \left( {\omega t - kx} \right)y=asin(ωt−kx)

Wave velocity v=ωk=25ππ/2=50 m/sec⁡v = {\omega \over k} = {{25\pi } \over {\pi /2}} = 50\,m/\sec v=kω​=π/225π​=50m/sec

The velocity of particle

vp=∂y∂t=75πcos⁡(25πt−π2x){v_p} = {{\partial y} \over {\partial t}} = 75\pi \cos \left( {25\pi t - {\pi \over 2}x} \right)vp​=∂t∂y​=75πcos(25πt−2π​x)

vpmax⁡=75π{v_{p_{\max}}} = 75\pi vpmax​​=75π

then vpmax⁡v=75π50=3π2{{{v_{{p_{\max }}}}} \over v} = {{75\pi } \over {50}} = {{3\pi } \over 2}vvpmax​​​=5075π​=23π​

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