datavorous/entrance-exam-dataset huggingface dataset:
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is _____ s

Options:

The correct answer is:
1

Solution:
Given here, A=8 cm and T=6 s

The displacement equation in SHM, when amplitude is halved is Acos2πtT=A2

⇒2πtT=π3

Time taken by the particle is t= T6=66=1sec.

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (27 Jun Shift 2)
A particle executes simple harmonic motion represented by displacement function as x(t)=Asin(ωt+ϕ). If the position and velocity of the particle at t=0 s are 2 cm and 2ω cm s-1 respectively, then its amplitude is x2 cm where the value of x is

Options:

The correct answer is:
2

Solution:
x(t)=Asin(ωt+ϕ)

v(t)=Aωcos(ωt+ϕ)

2=Asinϕ ...1

2ω=Aωcosϕ ...2

From (1) and (2)

tanϕ=1

ϕ=45°

Putting the value of ϕ in equation (1)

2=A12

A=22

x=2

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (27 Jul Shift 2)
A particle executes simple harmonic motion with a time period of 16 s along the x - axis. At time t=2 s, the particle crosses the mean position while at t=4 s, the velocity of the particle is + 4 m s-1. The amplitude of motion (in m) is

Options:
A 2π
B 162π
C 242π
D 322π

The correct answer is:
322π

Solution:
The general equation of SHM is

x=Asinωt+ϕ, here

ω=2π16=π8 rad s-1

At t=2 s, the particle is at the mean position. Hence,

0=Asinπ8×2+ϕ

⇒ϕ=-π4

So the equation of SHM will become

x=Asinπ8t-π4

v=πA8cosπ8t-π4

At t= 4 s

4=πA8cosπ8×4-π4

4=πA8cosπ4

⇒A=322π m

Topic: Physics, Oscillations, JEE Main
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm s-1. The distance of the particle from the mean position when its speed becomes 5 cm s-1 is α cm, where α=______.

Options:

The correct answer is:
12

Solution:
The formula to calculate the velocity of the particle at the mean position is given by

V=Aω ...1

From equation (1), it follows that

10=4×ω⇒ω=52 rad s-1

The formula to calculate the velocity of the particle at any instantaneous position x can be expressed as

v=ωA2-x2 ...2

From equation (2), it follows that

5=5242-x2⇒x2=16-4=12⇒x=12 cm

Hence, α=12.

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (27 Jan Shift 1)
A particle executes simple harmonic motion with an amplitude of 5 cm . When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:

Options:
A 8π3
B 38π
C 4π3
D 73π

The correct answer is:
8π3

Solution:
v=ω52-42=3ω
a=ω2×4
a=v
4ω2=3ω
ω=34=2πT
T=8π3 sec

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A particle executing a simple harmonic motion has a period of $6 \mathrm{~s}$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

Options:
A $\frac{3}{2} \mathrm{~s}$
B $\frac{1}{2} \mathrm{~s}$
C $\frac{3}{4} \mathrm{~s}$
D $\frac{1}{4} \mathrm{~s}$

The correct answer is:
$\frac{1}{2} \mathrm{~s}$

Solution:
Equation for simple harmonic motion
$\begin{aligned} y &=a \sin \left(\frac{2 \pi}{T}\right) t \\ \frac{a}{2} &=a \sin \left(\frac{2 \pi}{T}\right) t \\ \frac{2 \pi}{T} t &=\frac{\pi}{6} \\ t &=\frac{T}{12}=\frac{6}{12}=\frac{1}{2} \mathrm{~s} \end{aligned}$

Topic: Physics, Oscillations, JEE Main
A particle executing SHM along a straight line has zero velocity at points \(A\) and \(B\) whose distance from \(O\) on the same line \(O A B\) are \(a\) and \(b\), respectively. If the velocity at the mid point between \(A\) and \(B\) is \(v\), then its time period is

Options:
A \(\frac{\pi(b+a)}{v}\)
B \(\pi\left(\frac{b-a}{v}\right)\)
C \(\left(\frac{b+a}{2 v}\right)\)
D \(\left(\frac{b-a}{2 v}\right)\)

The correct answer is:
\(\pi\left(\frac{b-a}{v}\right)\)

Solution:
According to the question,
\(\therefore\) Amplitude
\(=\frac{\text { Distance travelled by the particles }}{2}(A \text { to } B)\)
Amplitude of particles executing simple harmonic motion (SHM) along a straight line \(A B\) is \((a)=\frac{b-a}{2}\).
Velocity of particle, \(v=\) Amplitude \(\times\) Oscillation frequency
\(\begin{aligned}
\therefore \quad v & =a \omega \\
v & =\left(\frac{b-a}{2}\right) \omega \text { or } \omega=\frac{2 v}{b-a} \quad \ldots (i)
\end{aligned}\)
\(\therefore\) Time period, \(T=\frac{2 \pi}{\omega}\)
Putting the value of \(\omega\) from Eq. (i) to above formula,
\(\Rightarrow \quad T=\frac{2 \pi}{2 v} \times(b-a) \Rightarrow T=\frac{b-a}{v} \pi\)
So, the time period of a particle executing SHM along a straight line from points \(A\) to \(B\) is,
\(T=\frac{b-a}{v} \pi\)

Topic: Physics, Oscillations, JEE Main
A particle executing SHM has a maximum speed of 0.5 m s-1 and a maximum acceleration of 1.0 m s-2. The angular frequency of oscillation is

Options:
A 2 rad s-1
B 0.5 rad s-1
C 2π rad s-1
D 0.5π rad s-1

The correct answer is:
2 rad s-1

Solution:
In SHM,
Maximum velocity vmax=ωA
Maximum acceleration amax=Aω2
⇒ ω=Aω2Aω=amaxvmax
vmax=0.5 m s-1
amax=1.0 m s-1
ω=10.5=105 =2 rad s-1

Topic: Physics, Oscillations, JEE Main
A particle executing SHM has a maximum speed of \( 0.5 \mathrm{~ms}^{-1} \) and maximum acceleration of
\( 1.0 \mathrm{~ms}^{-2} \). The angular frequency of oscillation is

Options:
A \( 2 \operatorname{rad} \mathrm{s}^{-1} \)
B \( 0.5 \operatorname{rad} \mathrm{s}^{-1} \)
C \( 2 \Pi \mathrm{rad} \mathrm{s}^{-1} \)
D \( 0.5 \Pi \mathrm{rad} \mathrm{s}^{-1} \)

The correct answer is:
\( 2 \operatorname{rad} \mathrm{s}^{-1} \)

Solution:
Given, maximum speed, $v_{\max }=0.5 m s^{-1} ;$ maximum acceleration,$a_{\max }=1.0 m s^{-1} ;$ angular frequency, $\omega=?$
Now, we know $v_{\max }=\omega A \rightarrow(1)$
where A is amplitude of oscillation.
Also, $a_{\max }=\omega^{2} A \rightarrow(2)$
Using Eqs. (1) and (2), we get
$\frac{v_{\max }}{a_{\max }}=\frac{\omega A}{\omega^{2} A}$
$\Rightarrow \frac{v_{\max }}{a_{\max }}=\frac{1}{\omega}$
$\Rightarrow \omega=\frac{a_{\max }}{v_{\max }}=\frac{1.0}{0.5}$
$\Rightarrow \omega=2 \operatorname{rad} s^{-1}$
Therefore, angular frequency of oscillation is $2 \mathrm{rads}^{-1}$

Topic: Physics, Oscillations, JEE Main
A particle executing S.H.M. of amplitude $4 \mathrm{~cm}$ and $T=4 \mathrm{sec}$. The time taken by it to move from positive extreme position to half the amplitude is

Options:
A $1 \mathrm{sec}$
B $1 / 3 \mathrm{sec}$
C $2 / 3 \mathrm{sec}$
D $\sqrt{3 / 2} \sec$

The correct answer is:
$2 / 3 \mathrm{sec}$

Solution:
Equation of motion $y=a \cos \omega t$
$\begin{aligned} & \Rightarrow \frac{a}{2}=a \cos \omega t \Rightarrow \cos \omega t=\frac{1}{2} \Rightarrow \omega t=\frac{\pi}{3} \\ & \Rightarrow \frac{2 \pi t}{T}=\frac{\pi}{3} \Rightarrow t=\frac{\frac{\pi}{3} \times T}{2 \pi}=\frac{4}{3 \times 2}=\frac{2}{3} \mathrm{sec}\end{aligned}$

Topic: Physics, Oscillations, JEE Main
A particle executing simple harmonic motion of amplitude $5 \mathrm{~cm}$ has maximum speed of $31.4 \mathrm{~cm} / \mathrm{s}$. The frequency of its oscillation is:

Options:
A $4 \mathrm{~Hz}$
B $3 \mathrm{~Hz}$
C $2 \mathrm{~Hz}$
D $1 \mathrm{~Hz}$

The correct answer is:
$1 \mathrm{~Hz}$

Solution:
Here $a=5 \mathrm{~cm}, V_{\max }=\frac{31.4 \mathrm{~cm}}{\mathrm{~s}}$
$\begin{aligned}
& V_{\max }=\omega a=31.4=2 \pi \mathrm{v} \times 5 \\
& \Rightarrow 31.4=10 \times 3.14 \times V \\
& \Rightarrow \quad V=1 Hz
\end{aligned}$

Topic: Physics, Oscillations, JEE Main
A particle experiences a variable force F→=4xi^+3y2j^ in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point 1,2 to point 2,3 in the x-y plane, then Kinetic Energy changes by :

Options:
A 25 J
B 50 J
C 12.5 J
D 0 J

The correct answer is:
25 J

Solution:
Using Work energy theorem,

Kinetic energy ∆KE=W=∫124xdx+∫233y2dy

=4x2212+3y3323

=2×22-2×12+32-23=6+19

=25 Joules

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (24 Jun Shift 1)
A particle falls from a height $h$ upon a fixed horizontal plane and rebounds. If $e$ is the coefficient of restitution, the total distance travelled before rebounding has stopped is

Options:
A $h\left(\frac{1+e^2}{1-e^2}\right)$
B $h\left(\frac{1-e^2}{1+e^2}\right)$
C $\frac{h}{2}\left(\frac{1-e^2}{1+e^2}\right)$
D $\frac{h}{2}\left(\frac{1+e^2}{1-e^2}\right)$

The correct answer is:
$h\left(\frac{1+e^2}{1-e^2}\right)$

Solution:
$h\left(\frac{1+e^2}{1-e^2}\right)$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle gets displaced by $\Delta \bar{r}=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \mathrm{m}$ under the action of a force $\vec{F}=(7 \hat{i}+4 \hat{j}+3 \hat{k})$. The change in its kinetic energy is

Options:
A
$38 \mathrm{~J}$

B
$70 \mathrm{~J}$

C
$52.5 \mathrm{~J}$

D
$126 \mathrm{~J}$

The correct answer is:
$38 \mathrm{~J}$

Solution:
According to work-energy theorem, Change in kinetic energy $=$ work done $=\vec{F} \cdot \Delta \vec{r}=(7 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})$ $=14+12+12=38 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2012 (07 May Online)
A particle goes from A to B with a speed of 40 km/h and B to C with a speed of 60 km/h . If AB=6âBC , the average speed in km/h between A and C is

Options:

The correct answer is:
42

Solution:
V av = distance travelled time taken
⇒ V av = AB + BC AB 4 0 + BC 6 0
⇒ V av = 6BC + BC 6BC 4 0 + BC 6 0
V av = 7 × 2 4 0 3 6 + 4 ⇒ V av = 7 × 2 4 0 4 0
⇒ V av = 4 2 km/h

Topic: Physics, Motion In One Dimension, JEE Main
A particle has an initial velocity $3 \hat{i}+4 \hat{j}$ and an acceleration of $0.4 \hat{i}+0.3 \hat{j}$. Its speed after 10 s is

Options:
A
10 units

B
$7 \sqrt{2}$ units

C
7 units

D
$8.5$ units

The correct answer is:
$7 \sqrt{2}$ units

Solution:
$$
\begin{aligned}
\overrightarrow{\mathrm{u}} & =3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} ; \overrightarrow{\mathrm{a}}=0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}} \\
\overrightarrow{\mathrm{u}} & =\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{a}} \mathrm{t} \\
& =3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+(0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}}) 10=3 \hat{i}+\hat{\mathrm{j}}+4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}=7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}
\end{aligned}
$$
Speed is $\sqrt{7^2+7^2}=7 \sqrt{2}$ units

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2009
A particle has displacement of $12 \mathrm{~m}$ towards east and $5 \mathrm{~m}$ towards north then $6 \mathrm{~m}$ vertically upward. The sum of these displacements is

Options:
A 12 m
B $10.04 \mathrm{~m}$
C $14 .31 \mathrm{~m}$
D None of these

The correct answer is:
$14 .31 \mathrm{~m}$

Solution:
$R=\sqrt{12^2+5^2+6^2}=\sqrt{144+25+36}=\sqrt{205}=14.31 \mathrm{~m}$

Topic: Physics, Mathematics in Physics, JEE Main
A particle has simple harmonic motion. The equation of its motion is $\mathrm{x}=5 \sin (4 t-\pi / 6)$, where $\mathrm{x}$ is its displacement. If the displacement of the particle is 3 units, then its velocity is

Options:
A $\frac{2 \pi}{3}$
B $\frac{5 \pi}{6}$
C 16 units
D 20 units

The correct answer is:
16 units

Solution:
From the given equation,
The amplitude of the wave is $\mathrm{a}=5$ and the angular velocity of the wave is $\omega=4$
The velocity of the wave can be obtained as: $\therefore \mathrm{v}=\omega \sqrt{\mathrm{a}^2-\mathrm{y}^2}$ $=4 \sqrt{(5)^2-(3)^2}=16$ units

Topic: Physics, Oscillations, JEE Main
A particle has two velocities of equal magnitude inclined to each other at an angle $\theta$. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then $\theta$ is

Options:
A
$90^{\circ}$

B
$120^{\circ}$

C
$45^{\circ}$

D
$60^{\circ}$

The correct answer is:
$120^{\circ}$

Solution:
$\tan \frac{\theta}{4}=\frac{\frac{u}{2} \sin \theta}{u+\frac{u}{2} \cos \theta}$
$\Rightarrow \sin \frac{\theta}{4}+\frac{1}{2} \sin \frac{\theta}{4} \cos \theta=\frac{1}{2} \sin \theta \cos \frac{\theta}{4}$
$\therefore 2 \sin \frac{\theta}{4}=\sin \frac{3 \theta}{4}=3 \sin \frac{\theta}{4}-4 \sin ^3 \frac{\theta}{4}$
$\therefore \sin ^2 \frac{\theta}{4}=\frac{1}{4} \Rightarrow \frac{\theta}{4}=30^{\circ}$ or $\theta=120^{\circ}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2006
A particle having a mass of $10^{-2} \mathrm{~kg}$ carries a charge of $5 \times 10^{-8} \mathrm{C}$. The particle is given an initial horizontal velocity of $10^5 \mathrm{~ms}^{-1}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$. To keep the particle moving in a horizontal direction, it is necessary that
(1) $\vec{B}$ should be perpendicular to the direction of velocity and $\vec{E}$ should be along the direction of velocity.
(2) Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity.
(3) Both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity.
(4) $\vec{B}$ should be along the direction of velocity and $\vec{E}$ should be perpendicular to the direction of velocity.
Which one of the following pairs of statements is possible?

Options:
A (1) and (3)
B (3) and (4)
C (2) and (3)
D (2) and (4)

The correct answer is:
(2) and (3)

Solution:
Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity and both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity.

Topic: Physics, Electrostatics, NEET, NEET 2010 (Mains)
A particle having a mass of 1.0 mg has a velocity of 3600 km/h . Calculate the wavelength of the particle.

Options:
A 6.626Ã10-31 m
B 6.626Ã10-30 m
C 6.626Ã10-29 m
D 6.626Ã10-28 m

The correct answer is:
6.626Ã10-31 m

Solution:
v=3600km/hr=1000 m/s
p=mv=1Ã10-6kg1000m/s=10-3 kg m/s
Use de broglie's relation.
Î»= h/mv=6.626Ã10-34J-s1Ã10-3 kg
=6.626Ã10-31 m

Topic: Chemistry, Structure of Atom, JEE Main
A particle having almost zero mass and exactly zero charge is

Options:
A positron
B electron
C neutron
D neutrino

The correct answer is:
neutrino

Solution:
We know that the neutrino is the particle which has almost zero mass and exactly zero charge. Positron has the charge $+e$ and mass $m_{e}$ [equal to electron mass].
Electron is a particle which has $-e$ charge and $9.1 \times 10^{-31}$ kg mass.
Neutron is a particle which has zero charge and $1838 \mathrm{~m}_{e}$ mass.
So, the required answer is neutrino.

Topic: Physics, Nuclear Physics, JEE Main
A particle having mass m and change q is released from the origin in a region in which electric field and magnetic fields are given by B→=-B0j^ and E→=E0k^. The speed of the particle as a function of its z-coordinate is

Options:
A 2qVB0+qE0Zm
B -qVB0+qE02Zm
C qE0Z2m
D 2qE0Zm

The correct answer is:
2qE0Zm

Solution:
mV22=qE0Z [As magnetic force does zero work]

v=2qE0Zm

Topic: Physics, Magnetic Effects of Current, JEE Main
A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron =1.6×10-19C )

Options:
A 9.1Ã10-31 kg
B 1.6Ã10-27Â kg
C 2.0Ã10-24 kg
D 1.6Ã10-19Â kg

The correct answer is:
2.0Ã10-24 kg

Solution:
Force to due to magnetic field
F=evB=mv2R
with electric field also, net force is zero.
∴eE=evB
⇒E=vB
E=eBRm. B
⇒m=eB2RE=1.6×10-19×0.52 . 0.5×10-2100 kg
=2.0×10-24 kg

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A particle initially at rest starts moving from reference point x=0 along x-axis, with velocity v that varies as v=4x m s-1. The acceleration of the particle is _____ m s-2.

Options:

The correct answer is:
8

Solution:
Given: v=4x

Instantaneous acceleration of the particle can be written as, a=vdvdx

=4x×4×12x-1/2=8 m s-2

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A particle is acted upon by a constant power. Then, which of the following physical quantity remains constant?

Options:
A Speed
B Rate of change of acceleration
C Kinetic energy
D Rate of change of kinetic energy

The correct answer is:
Rate of change of kinetic energy

Solution:
We have Power $=$ Rate of doing work $=\frac{d W}{d t}$
or, $\quad P=\frac{d W}{d t}=$ rate of change in $\mathrm{KE}$
i.e., $\quad P=\frac{d W}{d t}=\frac{d(\mathrm{KE})}{d t}=$ constant

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2013
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that

Options:
A
its velocity is constant

B
its acceleration is constant

C
its kinetic energy is constant

D
it moves in a straight line.

The correct answer is:
its kinetic energy is constant

Solution:
When a force of constant magnitude which is always perpendicular to the velocity of the particle acts on a particle, the work done and hence change in kinetic energy is zero.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2004
A particle is acted upon by constant forces $4 I+J-3 k$ and $3 I+J-k$ which displace it from a point $\hat{i}+2 \hat{j}+3 \hat{k}$ to the point $5 \hat{i}+4 \hat{j}+\hat{k}$. The work done in standard units by the forces is given by

Options:
A
40

B
30

C
25

D
15

The correct answer is:
40

Solution:
Work done by the forces $\vec{F}_1$ and $\vec{F}_2$ is $\left(\vec{F}_1+\vec{F}_2\right) \cdot \vec{d}$, where $\vec{d}$ is displacement According to question $\vec{F}_1+\vec{F}_2=(4 \hat{i}+\hat{j}-3 \hat{k})+(3 \hat{i}+\hat{j}-\hat{k})=7 \hat{i}+2 \hat{j}-4 \hat{k}$ and $\vec{d}=(5 \hat{i}+4 \hat{j}+\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=4 \hat{i}+2 \hat{j}-2 \hat{k}$. Hence $\left(\vec{F}_1+\vec{F}_2\right) \cdot \vec{d}$ is 40

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
A particle is attached to a string and the other end of the string is attached to a fixed support. In order to just complete the vertical circular motion, what should be the ratio of the kinetic energy of the particle at its highest point to that at its lowest point?

Options:

The correct answer is:
0.2

Solution:
K.E =12 mv2
At lowest point in vertical circular motion Vl=5rg and at highest point Vh=rg
â´ K.EhK.El=15=0.2

Topic: Physics, Work Power Energy, JEE Main
A particle is doing simple harmonic motion of amplitude $0.06 \mathrm{~m}$ and time period $3.14 \mathrm{~s}$. The maximum velocity of the particle is _______ $\mathrm{cm} / \mathrm{s}$.

Options:

The correct answer is:
12

Solution:
We know
$\begin{aligned}
\mathrm{V}_{\max } & =\omega \mathrm{A} \quad \text { at mean position } \\
& =\frac{2 \pi}{\mathrm{T}} \mathrm{A}=\frac{2 \pi}{\pi} \times 0.06=0.12 \mathrm{~m} / \mathrm{sec} \\
\mathrm{V}_{\max } & =12 \mathrm{~cm} / \mathrm{sec}
\end{aligned}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (06 Apr Shift 1)
A particle is executing S. H. M. of periodic time \( T \). The time taken by a particle in moving from mean position to half the maximum displacement is \( \left(\sin 30^{\circ}=0.5\right) \)

Options:
A \( \frac{T}{2} \)
B \( \frac{T}{4} \)
C \( \frac{\frac{r}{s}}{\frac{1}{s}} \)
D \( \frac{T}{12} \)

The correct answer is:
\( \frac{T}{12} \)

Solution:
The equation of displacement of SHM in terms of amplitude and angular frequency is

x=Asinωt

Here equation is given,

x=A2

⇒A2=Asinωt

⇒ωt=π6

⇒2πTt=π6

⇒t=T12

Topic: Physics, Oscillations, JEE Main
A particle is executing SHM along a straight line. Its velocities at distances \( \mathrm{x}_{1} \) and \( \mathrm{x}_{2} \) from the mean position are \( \mathrm{V}_{1} \) and \( \mathrm{V}_{2} \)
respectively. Its time period is:

Options:
A \( 2 \pi \sqrt{\frac{x_{1}^{2}+x_{2}^{2}}{V_{1}^{2}+V_{2}^{2}}} \)
B \( 2 \pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}} \)
C \( 2 \pi \sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{V}_{2}^{2}}{\mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}}} \)
D \( 2 \pi \sqrt{\frac{\mathrm{v}_{1}^{2}-\mathrm{V}_{2}^{2}}{\mathrm{x}_{1}^{2}-\mathrm{x}_{2}^{2}}} \)

The correct answer is:
\( 2 \pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}} \)

Solution:
For particle undergoing SHM,
V=ÏA2-x2âV2=Ï2 A2-x2
V1=ÏA2-x12âV12=Ï2A2-x12 ...(i)
V2=ÏA2-x22 âV22=Ï2A2-x22 ...(ii)
V12-V22=Ï2x22-x12
Ï=V12-V22x22-x12
T=2Ïx22-x12V12-V22

Topic: Physics, Oscillations, JEE Main
A particle is executing simple harmonic motion (SHM) of amplitude A, along the x -axis, about x=0. When its potential Energy PE equal kinetic energy KE, the position of the particle will be:

Options:
A A
B A2
C A22
D A2

The correct answer is:
A2

Solution:
PE=12kx2& KE=12KA2-x2
As given in question
KE=PE
⇒12KA2-x2=12Kx2
⇒2x2=A2
⇒x=±A2

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

Options:
A 1 : 1
B 1 : 3
C 2 : 1
D 1 : 4

The correct answer is:
1 : 3

Solution:
The formula to calculate the potential energy of the particle executing SHM is given by

P=12mω2x2 ...1

The formula to calculate the kinetic energy of the same particle is given by

K=12mω2A2-x2 ...2

Divide equation (1) by equation (2) to obtain the required ratio.

PK=12mω2x212mω2A2-x2=x2A2-x2 ...3

Substitute A2 for x into equation (3) to calculate the required ratio.

PK=A22A2-A22= 13

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A particle is executing simple harmonic motion with a time period T. At time t=0, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like:

Options:
A

B

C

D

The correct answer is:

Solution:
x=Asin⁡ω t
v=dxdt=a ωcos⁡ω t
K.E.=12m v2
=12 m A2ω2cos2⁡ ωt
=12 m A2ω2cos2⁡2πT t
At t=0, K.E. is maximum.
The period of K.E. graph is half of the period of displacement graph.
Comment:- The points on the graph of x-axis should be differentiable.
In all option mentioned, the graph has cusps at the points on x-axis.

Topic: Physics, Oscillations, JEE Main, JEE Main 2017 (02 Apr)
A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particles
after \( 2 \mathrm{~T} \) period from its initial position is

Options:
A A
B \( 4 \mathrm{~A} \)
C \( 8 \mathrm{~A} \)
D Zero

The correct answer is:
Zero

Solution:
It is the least interval of time after which the periodic motion of a body repeats itself. Therefore, displacement will be zero.

Topic: Physics, Oscillations, JEE Main
A particle is executing uniform circular motion with velocity $\vec{v}$ and acceleration $\vec{a}$. Which of the following is true?

Options:
A $\vec{v}$ is a constant; $\vec{a}$ is a constant
B $\vec{v}$ is not a constant; $\vec{a}$ is a constant
C $\vec{v}$ is a constant; $\vec{a}$ is not a constant
D $\vec{v}$ is not a constant; $\vec{a}$ is not a constant

The correct answer is:
$\vec{v}$ is not a constant; $\vec{a}$ is not a constant

Solution:
$\vec{v}$ is not a constant; $\vec{a}$ is not a constant
In uniform circular motion $\vec{v}$ keeps changing at every instant while magnitude of a remains same but its direction keeps changing.
Thus $\vec{a}$ is also not constant.

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2023 (Manipur)
A particle is given a horizontal velocity u, from the point P on a smooth horizontal floor which has a vertical circular track at the end. ABC is a semicircular track of radius r in the vertical plane. If the path length PA is x=3r and the particle returns to point P, then the initial speed of the particle is

Options:
A u=5gr
B u=52gr
C u=4gr
D u=32gr

The correct answer is:
u=52gr

Solution:
vC2=u2-2gÃ2r
vC2=u2-4gr â¦ 1
x=vC4rg â¦ 2
or, x2=vC24rg
as, x=3r
â´ 9r2=vC24rg
vC2=94 gr
From equation (1) u2-4gr=94gr
âu2=254gr or u=52gr

Topic: Physics, Laws of Motion, JEE Main