Q21JEE Main 2024NAT

A particle is doing simple harmonic motion of amplitude $0.06 {\text{m}}$ and time period $3.14 {\ s}$ . The maximum velocity of the particle is ______ ${\text{cm}} / {\ s}$ .

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📖 Explanation

We know ${\text{V}}_{\max }\;=\omega {\text{A}} \;\; \;\text{ at mean position }\;$ $\;=\;{2 \pi }/{ {\text{T}}} {\text{A}}=\;\frac{2 \pi }{\pi } \times 0.06=0.12 {\text{m}} / {\ sec}$ ${\text{V}}_{\max }\;=12 {\text{cm}} / {\ sec}$

Q22JEE Main 2024MCQ

The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10 {\text{m}}$ . If it dissipates $10 \%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is : $[Use, ${\text{g}}: 10 {\text{ms}}^{-2}$ ]$ [29-Jan-2024 Shift 2]

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📖 Explanation

$ℓ=10 {\text{m}}$ Initial energy $= {\text{mg}} ℓ$ $\;\;\text{ So, }\; \;\frac{9}{10} {\text{mg}} ℓ=\;\frac{1}{2} {\ mv}^2$ $\;\Rightarrow \;\frac{9}{10} \times 10 \times 10=\;\frac{1}{2} {\text{v}}^2$ $\; {\text{v}}^2=180$ $\; {\text{v}}={\sqrt{180}}=6 {\sqrt{5}} {\text{m}} / {\ s}$

Q23JEE Main 2024MCQ

A simple pendulum doing small oscillations at a place $R$ height above earth surface has time period of $T_1=4 {\ s} . {\text{T}}_2$ would be it's time period if is brought to a point which is at a height $2 R$ from earth surface. Choose the correct relation $[R=$ radius of Earth]:

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📖 Explanation

$\; {\text{T}}_1=2 \pi \sqrt{\;{ℓ}/{ {\text{GM}}}(2 {\text{R}})^2}$ $\; {\text{T}}_2=2 \pi \sqrt{\;{ℓ}/{ {\text{GM}}}(3 {\text{R}})^2}$ $\; \therefore \;{ {\text{T}}_1}/{ {\text{T}}_2}=\;\frac{2}{3}$

Q24JEE Main 2024NAT

A particle executes simple harmonic motion with an amplitude of $4 {\text{cm}}$ . At the mean position, velocity of the particle is $10 {\text{cm}} / {\ s}$ . The distance of the particle from the mean position when its speed becomes $5 {\text{cm}} / {\ s}$ is ${\sqrt{\alpha }} {\text{cm}}$ , where $\alpha =$ ___ [27-Jan-2024 Shift 1]

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📖 Explanation

$\; {\text{V}}_{\;\text{at mean position }\;}= {\text{A}} \omega \Rightarrow 10=4 \omega$ $\;\omega =\;\frac{5}{2}$ $\; {\text{V}}=\omega \sqrt{ {\text{A}}^{2-} {\text{x}}^2}$ $\;5=\;\frac{5}{2} \sqrt{4^{2-} {\text{x}}^2} \Rightarrow {\text{x}}^2=16-4$ $\; {\text{x}}={\sqrt{12}} {\text{cm}}$

Q25JEE Main 2024NAT

A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is $\;{2 {\text{A}}}/{3}$ . The new amplitude of motion is $\;\frac{n A}{3}$ . The value of $n$ is___ [31-Jan-2024 Shift 1]

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📖 Explanation

$\; {\text{v}}=\omega \sqrt{ {\text{A}}^{2-} {\text{x}}^2}$ $\;\;\text{ at }\; {\text{x}}=\;{2 {\text{A}}}/{3}$ $\; {\text{v}}=\omega \sqrt{ {\text{A}}^{2-} (\;{2 {\text{A}}}/{3}) ^2}=\;{{\sqrt{5}} {\text{A}} \omega }/{3}$ New amplitude $= {\text{A}}^{'}$ ${\text{v}}^{'}=3 {\text{v}}={\sqrt{5}} {\text{A}} \omega$ $=\omega \sqrt{ ( {\text{A}}^{'}) ^{2-} (\;{2 {\text{A}}}/{3}) ^2}$ ${\text{A}}^{'}=\;{7 {\text{A}}}/{3}$

Q26JEE Main 2024NAT

A mass $m$ is suspended from a spring of negligible mass and the system oscillates with a frequency $f_1$ . The frequency of oscillations if a mass $9 {\text{m}}$ is suspended from the same spring is $f_2$ . The value of $\;\frac{f_1}{f_2}$ is [1-Feb-2024 Shift 2]

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📖 Explanation

$\; {\text{f}}_1=\;\frac{1}{2 \pi } \sqrt{\;{ {\text{k}}}/{ {\text{m}}}}$ $\; {\text{f}}_2=\;\frac{1}{2 \pi } \sqrt{\;{ {\text{k}}}/{9 {\text{m}}}}$ $\;\;{ {\text{f}}_1}/{ {\text{f}}_2}=\sqrt{\;\frac{9}{1}}=\;\frac{3}{1}$

Q27JEE Main 2024NAT

A particle of mass $0.50 {\text{kg}}$ executes simple harmonic motion under force ${\text{F}}=-50 ( {\text{Nm}}^{-1}) {\text{x}}$ . The time period of oscillation is $\;\frac{x}{35} {\ s}$ . The value of ${\text{x}}$ is ______ . (Given $\pi =\;\frac{22}{7}$ )

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📖 Explanation

$\; {\text{m}}=0.5 {\text{kg}}$ $\; {\text{F}}=-50( {\text{x}})$ $\; {\text{ma}}=(-50 {\text{x}})$ $\;0.5 {\text{a}}=-50 {\text{x}}$ $\; {\text{a}}=(-100 {\text{x}})$ $\; {\text{W}}^2=100 \Rightarrow ( {\text{w}}=10)$ $\; {\text{T}}=\;\frac{2 \pi }{10}= (\;\frac{\pi }{5}) =\;\frac{22}{7 \times 15}= (\;\frac{22}{35})$ $\;\;\frac{\pi }{35}=\;\frac{22}{35} \Rightarrow {\text{x}}=22$

Q28JEE Main 2024NAT

The time period of simple harmonic motion of mass ${\text{M}}$ in the given figure is $\pi \sqrt{\;\frac{\alpha M}{5 K}}$ , where the value of $\alpha$ is _____ [31-Jan-2024 Shift 2]

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📖 Explanation

${\text{k}}_{ {\text{eq}}}=\;{2 {\text{k}} \cdot {\text{k}}}/{3 {\text{k}}}+ {\text{k}}=\;{5 {\text{k}}}/{3}$ Angular frequency of oscillation $(\omega )=\sqrt{\;{ {\text{k}}_{ {\text{eq}}}}/{ {\text{m}}}}$ $(\omega )=\sqrt{\;{5 {\text{k}}}/{3 {\text{m}}}}$ Period of oscillation $(\tau )=\;\frac{2 \pi }{\omega }=2 \pi \sqrt{\;{3 {\text{m}}}/{5 {\text{k}}}}$ $=\pi \sqrt{\;{12 {\text{m}}}/{5 {\text{k}}}}$

Q29JEE Main 2024MCQ

A simple pendulum of length $1 {\text{m}}$ has a wooden bob of mass $1 {\text{kg}}$ . It is struck by a bullet of mass $10^{-2} {\text{kg}}$ moving with a speed of $2 \times 10^2 {\text{ms}}^{-1}$ . The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use ${\text{g}}=10 {\text{m}} / {\ s}^2$ ) [1-Feb-2024 Shift 1]

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📖 Explanation

$\; {\mu}=( {\text{M}}+ {\text{m}}) {\text{V}}$ $\;10^{-2} \times 2 \times 10^2 ≅ 1 \times {\text{V}}$ $\; {\text{V}} ≅ 2 {\text{m}} / {\ s}$ $\; {\text{h}}=\;{ {\text{V}}^2}/{2 {\text{g}}}=0.2 {\text{m}}$

Q30JEE Main 2024NAT

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\;\frac{x}{8}$ , where ${\text{x}}=$ ____ [29-Jan-2024 Shift 1]

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📖 Explanation

Let total energy $= {\text{E}}=\;\frac{1}{2} {\text{KA}}^2$ $\; {\text{U}}=\;\frac{1}{2} {\text{K}} (\;{ {\text{A}}}/{3}) ^2=\;{ {\text{KA}}^2}/{2 \times 9}=\;{ {\text{E}}}/{9}$ $\; {\text{KE}}= {\text{E}}-\;{ {\text{E}}}/{9}=\;{8 {\text{E}}}/{9}$ $\;\;\text{ Ratio }\; \;{\;\text{ Total }\;}/{ {\text{KE}}}=\;{ {\text{E}}}/{{8 {\text{E}}}/{9}}=\;\frac{9}{8}$ $\; {\text{X}}=9$

Q31JEE Main 2024MCQ

In simple harmonic motion, the total mechanical energy of given system is ${\text{E}}$ . If mass of oscillating particle $P$ is doubled then the new energy of the system for same amplitude is :

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📖 Explanation

T.E. $=\;\frac{1}{2} {\text{kA}}^2$ since A is same T.E. will be same correct option (2)

Q32JEE Main 2024NAT

An object of mass $0.2 {\text{kg}}$ executes simple harmonic motion along ${\text{x}}$ axis with frequency of $(\;\frac{25}{\pi }) {\text{Hz}}$ . At the position ${\text{x}}=0.04 {\text{m}}$ the object has kinetic energy $0.5 {\text{J}}$ and potential energy $0.4 {\text{J}}$ . The amplitude of oscillation is ______ ${\text{cm}}$ .

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📖 Explanation

$\;\;\text{ Total energy }\;=\;\text{ K.E. }\;+ \;\text{ P.E. }\;$ $\;\;\text{ at }\; {\text{x}}=0.04 {\text{m}}, \;\text{ T.E. }\;=0.5+0.4=0.9 {\text{J}}$ $\;\text{ T.E }\;\;=1 {\text{m}} \omega ^2 {\text{A}}^2=0.9$ $\;=\;\frac{1}{2} \times 0.2 (2 \pi \times \;\frac{25}{\pi }) ^2 \times {\text{A}}^2=0.9$ $\Rightarrow {\text{A}}\;=0.06 {\text{m}}$ ${\text{A}}\;=6 {\text{cm}}$

Q33JEE Main 2024NAT

The displacement of a particle executing SHM is given by $x=10 sin \; (\omega t+\;\frac{\pi }{3}) {\text{m}}$ . The time period of motion is $3.14 {\ s}$ . The velocity of the particle at ${\text{t}}=0$ is _______ ${\text{m}} / {\ s}$ .

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📖 Explanation

Given, $\; {\text{T}}=3.14=\;\frac{2 \pi }{\omega }$ $\;\omega =2 {\text{rad}} / {\ s}$ $\; {\text{x}}=10 sin \; (\omega {\text{t}}+\;\frac{\pi }{3})$ $\; {\text{v}}=\;{ {\text{dx}}}/{ {\text{dt}}}=10 \omega \cos (\omega {\text{t}}+\;\frac{\pi }{3})$ $\;\;\text{ at }\; {\text{t}}=0$ $\; {\text{V}}=10 \omega \cos (\;\frac{\pi }{3}) =10 \times 2 \times \;\frac{1}{2}[\;\text{ using }\; \omega =2 {\text{rad}} / {\ s}]$ $\; {\text{v}}=10 {\text{m}} / {\ s}$

Q34JEE Main 2023NAT

A particle of mass $250 {\text{g}}$ executes a simple harmonic motion under a periodic force $F=(-25$ x) N. The particle attains a maximum speed of $4 {\text{m}} / {\ s}$ during its oscillation. The amplitude of the motion is ________ ${\text{cm}}$ . [29-Jan-2023 Shift 2]

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📖 Explanation

$\;\;\frac{1}{4} {\text{a}}=-25 {\text{x}} ; {\text{a}}=-100 {\text{x}}$ $\;\omega ^2=100 \;\; \omega =10,$ $\omega {\text{A}}=4 \;\; {\text{A}}=\;\frac{4}{10}=0.4 {\text{m}}$ ${\text{A}}=40 {\text{cm}}$

Q35JEE Main 2023NAT

In the figure given below. a block of mass ${\text{M}}=490 {\text{g}}$ placed on a frictionless table is connected with two springs having same spring constant $( {\text{K}}=2 {\text{N}} {\text{m}}^{-1})$ . If the block is horizontally displaced through ' ${\text{X}}$ 'm then the number of complete oscillations it will make in $14 \pi$ seconds will be _______ [31-Jan-2023 Shift 1]

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📖 Explanation

$\text{Keff}=\text{K}+\text{K}$ as both springs are in use in parallel $=2 {\text{k}}\;$ $=2 \times 2=4 {\text{N}} / {\text{m}} \;\; {\text{m}}\;=490 {\text{gm}}$ $\;=0.49 {\text{kg}}$ $T\;=2 \pi \sqrt{\;{ {\text{m}}}/{ {\text{Keff}}}}=2 \pi \sqrt{\;{0.49 {\text{kg}}}/{4}}$ $\;=2 \pi \sqrt{\;\frac{49}{400}}=2 \pi \;\frac{7}{20}=\;\frac{7 \pi }{10}$ No. of oscillation in the $14 \pi$ is ${\text{N}}=\;{\;\text{ time }\;}/{ {\text{T}}}=\;\frac{14 \pi }{7 \pi / 10}=20$ Ans in 20 .

Q36JEE Main 2023MCQ

The variation of kinetic energy $( {\text{KE}})$ of a particle executing simple harmonic motion with the displacement $( {\text{x}})$ starting from mean position to extreme position (A) is given by [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{K}} \cdot {\text{E}}= {\text{T}} \cdot {\text{E}}- {\text{P}} \cdot {\text{E}}$ $\; {\text{K}} \cdot {\text{E}}=\;\frac{1}{2} {\text{KA}} 2-\;\frac{1}{2} {\text{Kx}}^2$ Graph ${\text{b}} / {\text{w}} {\text{K}} \cdot {\text{E}}$ and ${\text{x}}$ will be parabola Option $\to$ (1)

Q37JEE Main 2023NAT

The velocity of a particle executing SHM varies with displacement $( {\text{x}})$ as $4 {\text{v}}^2=50- {\text{x}}^2$ . The time period of oscillations is $\;\frac{x}{7} {\ s}$ . The value of $x$ is _______. $(\;\text{ Take }\; \pi =\;\frac{22}{7})$ [30-Jan-2023 Shift 2]

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📖 Explanation

$\;4 {\text{v}}^2=50- {\text{x}}^2$ $\;\Rightarrow \;\; {\text{v}}=\;\frac{1}{2} \sqrt{50- {\text{x}}^2}$ $\;\omega =\;\frac{1}{2}$ $\; {\text{T}}=\;\frac{2 \pi }{\omega }=4 \pi =\;\frac{88}{7}$ $\; {\text{x}}=88$

Q38JEE Main 2023MCQ

Which graph represents the difference between total energy and potential energy of a particle executing SHM vs it's distance from mean position? [13-Apr-2023 shift 1]

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📖 Explanation

Total energy in ${\text{SHM}}= {\text{E}}$ $\;E=K+U$ $\;E-U=K$ $\;E-U=\;\frac{1}{2} m \omega ^2 (A^{2-}x^2)$

Q39JEE Main 2023NAT

A simple pendulum with length $100 {\text{cm}}$ and bob of mass $250 {\text{g}}$ is executing S.H.M. of amplitude $10 {\text{cm}}$ . The maximum tension in the string is found to be $\;\frac{x}{40} {\text{N}}$ . The value of $x$ is _______. [6-Apr-2023 shift 2]

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📖 Explanation

For pendulum ${\text{T}}_{\max }= {\text{mg}}+\;{ {\ mv}^2}/{ {\text{L}}}$ . . . (1)Given ${\text{m}}=\;\frac{1}{4} {\text{kg}}, {\text{L}}=1 {\text{m}}, {\text{g}}=9.8 {\text{m}} / {\ s}^2$ and amplitude ${\text{A}}=\;\frac{1}{10} {\text{m}}$ For SHM, ${\text{KE}}_{\max }=\;\frac{1}{2} {\ mv}^2=\;\frac{1}{2} {\text{m}} \omega ^2 {\text{A}}^2$ using $\omega =\sqrt{\;{ {\text{g}}}/{ {\text{L}}}}$ $m v^2=m (\sqrt{\;{ {\text{g}}}/{ {\text{L}}}}) ^2 {\text{A}}^2=\;{ {\text{mgA}}}/{ {\text{L}}}$ . . . (2)using (2) in (1) ${\text{T}}_{\max }=2 {\text{mg}}+\;{ {\text{mgA}}^2}/{ {\text{L}}^2}$ $=m g \$ [1+;\frac{1}{10^2}]$ =;\frac{1}{4} \times 9.8 \times ;\frac{101}{100} $or$ {\text{T}}_{\max }=;\frac{98.98}{40} $Therefore$ {\text{x}}=99$

Q40JEE Main 2023MCQ

Choose the correct length $( {\text{L}})$ versus square of time period $( {\text{T}}^2)$ graph for a simple pendulum executing simple harmonic motion. [1-Feb-2023 Shift 2]

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📖 Explanation

$\; {\text{T}}=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$ $\; {\text{T}}^2=\;{4 \pi ^2}/{ {\text{g}}} \times ℓ$ $\; {\text{T}}^2 \alpha ℓ$