- 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

Answer & Solution
(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
\({v} ={A} \sin \omega {t}+{Bcos} \omega {t}\)
\({{dt}}={A} \omega \cos \omega {t}-{B} \omega \sin \omega {t}\)
\({At} {t}=0, {x}(0)={B}\)
\({v}(0)={A} \omega\)
\({x}={A} \sin \omega {t}+{B} \sin \left(\omega {t}+90^{\circ}\right)\)
\(A_{\text {net }}=\sqrt{A^{2}+B^{2}}\)
\(\tan \alpha=\frac{B}{A} \Rightarrow \cot \alpha=\frac{A}{B}\)
\(\Rightarrow \quad x=\sqrt{A^{2}+B^{2}} \sin (\omega t+\alpha)\)
\(\Rightarrow \quad x=\sqrt{A^{2}+B^{2}} \cos (\omega t-(90-\alpha))\)
\(x=C \cos (\omega t-\phi)\)
\(\Rightarrow C=\sqrt{A^{2}+B^{2}}\)
\(C= \sqrt{\frac{[v(0)]^{2}}{\omega^{2}}+[x(0)]^{2}}\)
\(\phi= 90-\alpha\)
\(\tan \alpha=\cos \alpha=\frac{A}{B}\)
\(\Rightarrow \tan \phi=\frac{v(0)}{x(0) \cdot \omega}\)
\(\phi= \tan ^{-1}\left(\frac{v(0)}{x(0) \omega}\right)\)
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