The motion of a mass on a spring, with spring constant $K$ is as shown in figure. ![JEE Main 2021 Physics, Laws of Motion — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Laws_of_Motion/648b5a6a417cc3fb48d6714a/question_1__q_648b5a6a417cc3fb48d6714a__cdn-question-pool.getmarks.app__23e6b85d-f888-4c9e-bb35-d94ce908ca1c-image__54c1094254_final_ppt_sync.png) The equation of motion is given by, $x(t)=A\mathrm{sin}\omega t+$$B\mathrm{cos}\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 $x(t)=C\mathrm{cos}(\omega t-\phi ),$ where $C$ and $\phi$ are

$x=Asin\omega t+Bcos\omega t$ $v=\frac{dx}{dt}=A\omega \mathrm{cos}\omega t-B\omega \mathrm{sin}\omega t$ At $t=0,x(0)=B$ $v(0)=A\omega$ $x=A\mathrm{sin}(\omega +B\mathrm{sin}(\omega t+90^{\circ})$ ![JEE Main 2021 Physics, Laws of Motion — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Laws_of_Motion/648b5a6a417cc3fb48d6714a/solution_1__q_648b5a6a417cc3fb48d6714a__cdn-question-pool.getmarks.app__2b239dec-4435-440f-9650-43cf18df5b1c-image__d579f9478c_final_ppt_sync.png) ${A}_{\mathrm{net}}=\sqrt{{A}^{2}+{B}^{2}}$ $\mathrm{tan}\alpha =\frac{B}{A}\Rightarrow \mathrm{cot}\alpha =\frac{A}{B}$ $\Rightarrow x=\sqrt{{A}^{2}+{B}^{2}}\mathrm{sin}(\omega t+\alpha )$ $\Rightarrow x=\sqrt{{A}^{2}+{B}^{2}}\mathrm{cos}(\omega t-(90-\alpha ))$ $x=C\mathrm{cos}(\omega t-\phi )$ $\Rightarrow C=\sqrt{{A}^{2}+{B}^{2}}$ $C=\sqrt{\frac{[v(0){]}^{2}}{{\omega }^{2}}+[x(0){]}^{2}}$ $\phi =90-\alpha$ $\mathrm{tan}\alpha =\mathrm{cos}\alpha =\frac{A}{B}$ $\Rightarrow \mathrm{tan}\phi =\frac{v(0)}{x(0)\cdot \omega }$ $\phi ={\mathrm{tan}}^{-1}(\frac{v(0)}{x(0)\omega })$

JEE Main 2021 — Physics Mechanics

2021mcqhard

The motion of a mass on a spring, with spring constant $K$ is as shown in figure.

JEE Main 2021 Physics, Laws of Motion — question figure

The equation of motion is given by, $x(t)=A\mathrm{sin}\omega t+$ $B\mathrm{cos}\omega t$ 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 $x(t)=C\mathrm{cos}(\omega t-\phi ),$ where $C$ and $\phi$ are

Official previous-year question

Held on 22 Jul 2021 · Verified 6 Jul 2026.

Options

A
$C=\sqrt{\frac{2v(0{)}^{2}}{{\omega }^{2}}+x{(0)}^{2}},\phi ={\mathrm{tan}}^{-1}(\frac{v(0)}{x(0)\omega })$
B
$C=\sqrt{\frac{2v(0{)}^{2}}{{\omega }^{2}}+x(0{)}^{2}},\phi ={\mathrm{tan}}^{-1}(\frac{x(0)\omega }{2v(0)})$
C
$C=\sqrt{\frac{v(0{)}^{2}}{{\omega }^{2}}+x(0{)}^{2}},\phi ={\mathrm{tan}}^{-1}(\frac{x(0)\omega }{v(0)})$
D
$C=\sqrt{\frac{v(0{)}^{2}}{{\omega }^{2}}+x(0{)}^{2}},\phi ={\mathrm{tan}}^{-1}(\frac{v(0)}{x(0)\omega })$

Did you get this right?

Solution

$x=Asin\omega t+Bcos\omega t$

$v=\frac{dx}{dt}=A\omega \mathrm{cos}\omega t-B\omega \mathrm{sin}\omega t$

At $t=0,x(0)=B$

$v(0)=A\omega$

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

JEE Main 2021 Physics, Laws of Motion — solution figure

${A}_{\mathrm{net}}=\sqrt{{A}^{2}+{B}^{2}}$

$\mathrm{tan}\alpha =\frac{B}{A}\Rightarrow \mathrm{cot}\alpha =\frac{A}{B}$

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

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

$x=C\mathrm{cos}(\omega t-\phi )$

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

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

$\phi =90-\alpha$

$\mathrm{tan}\alpha =\mathrm{cos}\alpha =\frac{A}{B}$

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

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

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Options

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

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

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

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

Solution

$x=Asin\omega t+Bcos\omega t$

$v=\frac{dx}{dt}=A\omega \mathrm{cos}\omega t-B\omega \mathrm{sin}\omega t$

At $t=0,x(0)=B$

$v(0)=A\omega$

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

JEE Main 2021 Physics, Laws of Motion — solution figure

${A}_{\mathrm{net}}=\sqrt{{A}^{2}+{B}^{2}}$

$\mathrm{tan}\alpha =\frac{B}{A}\Rightarrow \mathrm{cot}\alpha =\frac{A}{B}$

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

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

$x=C\mathrm{cos}(\omega t-\phi )$

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

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

$\phi =90-\alpha$

$\mathrm{tan}\alpha =\mathrm{cos}\alpha =\frac{A}{B}$

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

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

JEE Main 2021 — Physics Mechanics

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JEE Main 2021 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics