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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2021
  • Hard
  • MCQ

The motion of a mass on a spring, with spring constant \({K}\) is as shown in figure. The equation of motion is given by \(x(t)= A sin \omega t+ Bcos\omega t\) with \(\omega=\sqrt{\frac{K}{m}}\) 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 \(x(t)=C \cos (\omega t-\phi)\), where \(C\) and \(\phi\) are

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Answer & Solution

Correct Answer

(D) \({C}=\sqrt{\frac{{v}(0)^{2}}{\omega^{2}}+{x}(0)^{2}}, \phi=\tan ^{-1}\left(\frac{{v}(0)}{{x}(0) \omega}\right)\)

Step-by-step Solution

Detailed explanation

\(x={A} \sin \omega {t}+{B} \sin \omega {t}\)

\({v} ={A} \sin \omega {t}+{Bcos} \omega {t}\)

\({{dt}}={A} \omega \cos \omega {t}-{B} \omega \sin \omega {t}\)

\({At} {t}=0, {x}(0)={B}\)

\({v}(0)={A} \omega\)

\({x}={A} \sin \omega {t}+{B} \sin \left(\omega {t}+90^{\circ}\right)\)

\(A_{\text {net }}=\sqrt{A^{2}+B^{2}}\)

\(\tan \alpha=\frac{B}{A} \Rightarrow \cot \alpha=\frac{A}{B}\)

\(\Rightarrow \quad x=\sqrt{A^{2}+B^{2}} \sin (\omega t+\alpha)\)

\(\Rightarrow \quad x=\sqrt{A^{2}+B^{2}} \cos (\omega t-(90-\alpha))\)

\(x=C \cos (\omega t-\phi)\)

\(\Rightarrow C=\sqrt{A^{2}+B^{2}}\)

\(C= \sqrt{\frac{[v(0)]^{2}}{\omega^{2}}+[x(0)]^{2}}\)

\(\phi= 90-\alpha\)

\(\tan \alpha=\cos \alpha=\frac{A}{B}\)

\(\Rightarrow \tan \phi=\frac{v(0)}{x(0) \cdot \omega}\)

\(\phi= \tan ^{-1}\left(\frac{v(0)}{x(0) \omega}\right)\)

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