Oscillations JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Oscillations, free to read — no sign-in needed. The full chapter has 30 questions; sign in to attempt the remaining 25 in the exam simulator.

1. Q1JEE Advanced Adv 2024 (Paper 1)
   A block of mass $5 \mathrm{~kg}$ moves along the $x$-direction subject to the force $F=(-20 x+10) \mathrm{N}$, with the value of $x$ in metre. At time $t=0 \mathrm{~s}$, it is at rest at position $x=1 \mathrm{~m}$. The position and momentum of the block at $t=(\pi / 4) \mathrm{s}$ are
   1. A.$-0.5 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
   2. B.$0.5 \mathrm{~m}, 0 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
   3. C.$0.5 \mathrm{~m},-5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
   4. D.$-1 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
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   Answer: (C)

   $\mathrm{F}=-20\left(\mathrm{x}-\frac{1}{2}\right)=-20 \mathrm{X} \quad\left(\mathrm{X}=\mathrm{x}-\frac{1}{2}\right)$ $\therefore$ Particle will perform SHM about $\mathrm{x}=\frac{1}{2}$ with $\omega=2 \mathrm{rad} / \mathrm{sec} \Rightarrow \mathrm{T}=\pi \mathrm{sec}$. $\therefore$ Phase covered in $t=\frac{\pi}{4}$ second $=90^{\circ}$. Given particle is at rest at $\mathrm{x}=1 \mathrm{~m} \Rightarrow \mathrm{x}=1$ is extreme position. $\therefore$ In $\frac{\pi}{4}$ sec, it will be at equilibrium $\therefore \mathrm{x}=0.5 \mathrm{~m}$ and momentum $=\mathrm{m} \omega \mathrm{A}=5 \times 2 \times 0.5=5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ Direction will be towards -ve $\mathrm{x}$. Hence option (3)
2. Q2JEE Advanced Adv 2022 (Paper 2)
   On a frictionless horizontal plane, a bob of mass $m=0.1kg$ is attached to a spring with natural length ${l}_{0}=0.1m$. The spring constant is ${k}_{1}=0.009N{m}^{-1}$ when the length of the spring $l\gt {l}_{0}$ and is ${k}_{2}=0.016N{m}^{-1}$ when $l\lt {l}_{0}$. Initially the bob is released from $l=0.15m$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=\left(n\pi \right)s$, then the integer closest to $n$ is _______.
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   Answer: 6

   Angular frequency for different values of spring constant will be given by, ${\omega }_{1}=\sqrt{\dfrac{{k}_{1}}{m}}$ and ${\omega }_{2}=\sqrt{\dfrac{{k}_{2}}{m}}$ Time period for a spring of fix spring constant, $k$ is given by, $=2\pi \sqrt{\dfrac{m}{k}}$. As per question, spring constant is having different values, hence we will have to find half of the time period for the given parts, first for the right half part of the oscillations & next for the left half part of the oscillation from the mean position. $∴$ Time period $=\pi \sqrt{\dfrac{m}{{k}_{1}}}+\pi \sqrt{\dfrac{m}{{k}_{2}}}$ $=\pi \sqrt{\dfrac{0.1}{0.009}}+\pi \sqrt{\dfrac{0.1}{0.016}}$ $=\dfrac{\pi }{0.3}+\dfrac{\pi }{0.4}$ $=\pi \times \left(\dfrac{4+3}{12}\right)\times 10$ $=\dfrac{70}{12}\pi$ $=5.83\pi$ Therefore, $n=6$.
3. Q3JEE Advanced Adv 2022 (Paper 2)
   A particle of mass $1kg$ is subjected to a force which depends on the position as $\vec{F}=-k\left(x\hat{i}+y\hat{j}\right)kgm{s}^{-2}$ with $k=1kg{s}^{-2}$. At time $t=0$, the particle's position $\vec{r}=\left(\dfrac{1}{\sqrt{2}}\hat{i}+\sqrt{2}\hat{j}\right)m$ and its velocity $\vec{v}=\left(-\sqrt{2}\hat{i}+\sqrt{2}\hat{j}+\dfrac{2}{\pi }\hat{k}\right)m{s}^{-1}$. Let ${v}_{x}$ and ${v}_{y}$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=0.5m$, the value of $\left(x{v}_{y}-y{v}_{x}\right)$ is ______ ${m}^{2}{s}^{-1}$.
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   Answer: 3

   Given here: $\vec{F}=-k\left(x\hat{i}+y\hat{j}\right)kgm{s}^{-2}$ and $m=1kg$ In x-direction, ${F}_{x}=-x=m{a}_{x}$ So, acceleration, ${a}_{x}=\dfrac{{d}^{2}x}{d{t}^{2}}=-x$ Now, for particle executing SHM, displacement along x-direction is $\Rightarrow x={A}_{x}\sin \left(\omega t+{\phi }_{x}\right)$, here, angular frequency, $\omega =1rad{s}^{-1}$ and velocity, ${v}_{x}={A}_{x}\omega \cos \left(\omega t+{\phi }_{x}\right)$ Given at $t=0,x=\dfrac{1}{\sqrt{2}}m$ and ${v}_{x}=-\sqrt{2}m{s}^{-1}$ So, putting the values, we get $\dfrac{1}{\sqrt{2}}={A}_{x}\sin {\phi }_{x}$ and $-\sqrt{2}={A}_{x}\cos {\phi }_{x}$ From above two equations, $\Rightarrow \tan {\phi }_{x}=-\dfrac{1}{2}...\left(1\right)$ And ${A}_{x}=\sqrt{\dfrac{5}{2}}m...\left(2\right)$Similarly, along y-direction, ${F}_{y}=-y=m{a}_{y}$, $\Rightarrow {a}_{y}=\dfrac{{d}^{2}y}{d{t}^{2}}=-y$ So, displacement, $y={A}_{y}\sin \left(\omega t+{\phi }_{y}\right)$ and velocity ${v}_{y}={A}_{y}\omega \cos \left(\omega t+{\phi }_{y}\right)$ Given at $t=0, y=\sqrt{2}m$ and ${v}_{y}=\sqrt{2}m{s}^{-1}$ So, putting the values, we get $\sqrt{2}={A}_{y}\sin {\phi }_{y}$ and $\sqrt{2}={A}_{y}\cos {\phi }_{y}$ From above two relations, we have $\Rightarrow {\phi }_{y}=\dfrac{\pi }{4}...\left(3\right)$ and ${A}_{y}=2m...\left(4\right)$ Now, the value of $\left(x{v}_{y}-y{v}_{x}\right)=\sqrt{\dfrac{5}{2}}\sin \left(\omega t+{\phi }_{x}\right)\times 2\cos \left(\omega t+{\phi }_{y}\right)-2\sin \left(\omega t+{\phi }_{y}\right)\times \sqrt{\dfrac{5}{2}}\cos \left(\omega t+{\phi }_{x}\right)$ $=\sqrt{\dfrac{5}{2}}\times 2\left(\sin \left(\omega t+{\phi }_{x}\right)\cos \left(\omega t+{\phi }_{y}\right)-\sin \left(\omega t+{\phi }_{y}\right)\times \cos \left(\omega t+{\phi }_{x}\right)\right.$ $=\sqrt{10}\sin \left({\phi }_{x}-{\phi }_{y}\right)$ $=\sqrt{10}\left(\sin {\phi }_{x}\cos {\phi }_{y}-\cos {\phi }_{x}\sin {\phi }_{y}\right)$ $=\sqrt{10}\left(\dfrac{1}{\sqrt{5}}\times \dfrac{1}{\sqrt{2}}-\left(-\dfrac{2}{\sqrt{5}}\right)\times \dfrac{1}{\sqrt{2}}\right)$ $=3$
4. Q4JEE Advanced Adv 2016 (Paper 2)
   A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position ${x}_{0}$. Consider two cases : (i) when the block is at ${x}_{0}\text{;}$ and (ii) when the block is at $x={x}_{0}+\text{A}\text{.}$ In both the cases, a particle with mass m $\left(\lt \text{M}\right)$ is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass m is placed on the mass M?
   1. A.The amplitude of oscillation in the first case changes by a factor of $\sqrt{\dfrac{\text{M}}{\text{m}+{\text{M}}^{'}}}$ whereas in the second case it remains unchanged
   2. B.The final time period of oscillation in both the cases is same
   3. C.The total energy decreases in both the cases
   4. D.The instantaneous speed at ${x}_{0}$ of the combined masses decreases in both the cases.
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   Answer: A,B,D

   ${\text{T}}_{\text{i}}=2\pi \sqrt{\dfrac{\text{M}}{\text{K}}},{\text{T}}_{\text{f}}=2\pi \sqrt{\dfrac{\text{M}+\text{m}}{\text{K}}}$ Case (i): $\text{M}\left(\text{A}\omega \right)=\left(\text{M}+\text{m}\right)\text{V}$ $∴$ Velocity decreases at equilibrium position. By energy conservation ${\text{A}}_{f}={\text{A}}_{\text{i}}\sqrt{\dfrac{\text{M}}{\text{M}+\text{m}}}$ Case (ii): No energy loss, amplitude remains same At equilibrium $\left({x}_{0}\right)velocity=\text{A}\omega$ In both cases $\omega$ decrease so velocity decreases in both cases
5. Q5JEE Advanced Adv 2010 (Paper 1)
   When two progressive waves $y_1=4 \sin (2 x-6 t) \quad$ and $y_2=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right) \quad$ are superimposed, the amplitude of the resultant wave is
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   Answer: 5

   Phase difference between the two waves is $90^{\circ}$. Amplitudes are added by vector method. $\therefore$ Answer is 5 .