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) ${A}_{\mathrm{net}}=\sqrt{{A}^{2}+{B}^{2}}$ $\mathrm{tan}\alpha =\frac{B}{A}\Rightarrow \mathrm{cot}\alpha =\frac{A}{B}$ $\Rightarrow x=\sqrt{{A}^{2}+{B}^{2}}\mathrm{sin}(\omega t+\alpha )$ $\Rightarrow x=\sqrt{{A}^{2}+{B}^{2}}\mathrm{cos}(\omega t-(90-\alpha ))$ $x=C\mathrm{cos}(\omega t-\phi )$ $\Rightarrow C=\sqrt{{A}^{2}+{B}^{2}}$ $C=\sqrt{\frac{[v(0){]}^{2}}{{\omega }^{2}}+[x(0){]}^{2}}$ $\phi =90-\alpha$ $\mathrm{tan}\alpha =\mathrm{cos}\alpha =\frac{A}{B}$ $\Rightarrow \mathrm{tan}\phi =\frac{v(0)}{x(0)\cdot \omega }$ $\phi ={\mathrm{tan}}^{-1}(\frac{v(0)}{x(0)\omega })$

The motion of a mass on a… - JEE Main 2021 Physics PYQ

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

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

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

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

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

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

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

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

$\phi =90-\alpha$

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

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

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

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Solution

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Options

Did you get this right?

Solution

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Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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