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Waves question

2021 · 22 Jul · Shift 2 · Q63
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  5. /2021 · 22 Jul · Shift 2 · Q63

Waves question

2021 · 22 Jul · Shift 2 · Q63

JEE MainPhysicsWavesMCQ+4 / −1
The motion of a mass on a spring, with spring constant K is as shown in figure. JEE Main 2021 (Online) 22th July Evening Shift Physics - Waves Question 64 English The equation of motion is given by x(t) = A sin ω\omegaω t + B cos ω\omegaω t with ω\omegaω=Km\sqrt {{K \over m}}mK​​ Suppose that at time t = 0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t) = C cos(ω\omegaω t −ϕ-\phi−ϕ), where C and ϕ\phiϕ are :
  1. A
    C=2v(0)2ω2+x(0)2,ϕ=tan⁡−1(x(0)ω2v(0))C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {2v(0)}}} \right)C=ω22v(0)2​+x(0)2​,ϕ=tan−1(2v(0)x(0)ω​)
  2. B
    C=v(0)2ω2+x(0)2,ϕ=tan⁡−1(x(0)ωv(0))C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {v(0)}}} \right)C=ω2v(0)2​+x(0)2​,ϕ=tan−1(v(0)x(0)ω​)
  3. C
    C=v(0)2ω2+x(0)2,ϕ=tan⁡−1(v(0)x(0)ω)C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)C=ω2v(0)2​+x(0)2​,ϕ=tan−1(x(0)ωv(0)​)
  4. D
    C=2v(0)2ω2+x(0)2,ϕ=tan⁡−1(v(0)x(0)ω)C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)C=ω22v(0)2​+x(0)2​,ϕ=tan−1(x(0)ωv(0)​)
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