(D) Given the equation $x = A \sin(\omega t + \phi)$.At $t = 0$,$x = A/2$,so $A/2 = A \sin(\phi)$,which gives $\sin(\phi) = 1/2$.This implies $\phi =

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(D) Given the equation $x = A \sin(\omega t + \phi)$.At $t = 0$,$x = A/2$,so $A/2 = A \sin(\phi)$,which gives $\sin(\phi) = 1/2$.This implies $\phi = \pi/6$ or $\phi = 5\pi/6$.The velocity of the particle is given by $v = dx/dt = A\omega \cos(\omega t + \phi)$.At $t = 0$,$v = A\omega \cos(\phi)$.Since the particle is moving in the negative $x$-direction,$v For $\phi = \pi/6$,$\cos(\pi/6) = \sqrt{3}/2 > 0$.For $\phi = 5\pi/6$,$\cos(5\pi/6) = -\sqrt{3}/2 Therefore,the correct phase angle is $\phi = 5\pi/6$.

Class 11 Physics Oscillations Position of a Particle in SHM, Displacement and Phase

Medium

The displacement-time equation of a particle executing $SHM$ is $x = A \sin(\omega t + \phi)$. At time $t = 0$,the position of the particle is $x = A/2$ and it is moving along the negative $x$-direction. Then the phase angle $\phi$ is:

A
$\pi / 6$
B
$\pi / 3$
C
$2\pi / 3$
D
$5\pi / 6$

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Oscillations

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Position of a Particle in SHM, Displacement and Phase

$A$ particle is executing $SHM$ of amplitude $A$ about the mean position $x = 0.$ Which of the following cannot be a possible phase difference between the positions of the particle at $x = +A/2$ and $x = -A/\sqrt{2}$ (in $^o$)?

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The acceleration of a particle starting from rest varies with time according to the relation $A = -a\omega^2 \sin\omega t$. The displacement of this particle at a time $t$ will be

Medium

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$A$ particle starts executing simple harmonic motion from one extreme position. If $a, b$ and $c$ are the displacements of the particle from the mean position at the ends of three successive seconds,the frequency of simple harmonic motion is

EasyAP EAMCET 2018

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$A$ particle is performing simple harmonic motion along the $x$-axis with an amplitude of $4 \, cm$ and a time period of $1.2 \, s$. The minimum time taken by the particle to move from $x = 2 \, cm$ to $x = +4 \, cm$ and back again is given by .... $s$.

DifficultAIIMS 1995

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$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

MediumMHT CET 2025

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The displacement-time equation of a particle executing $SHM$ is $x = A \sin(\omega t + \phi)$. At time $t = 0$,the position of the particle is $x = A/2$ and it is moving along the negative $x$-direction. Then the phase angle $\phi$ is:

Similar Questions

$A$ particle is executing $SHM$ of amplitude $A$ about the mean position $x = 0.$ Which of the following cannot be a possible phase difference between the positions of the particle at $x = +A/2$ and $x = -A/\sqrt{2}$ (in $^o$)?

The acceleration of a particle starting from rest varies with time according to the relation $A = -a\omega^2 \sin\omega t$. The displacement of this particle at a time $t$ will be

$A$ particle starts executing simple harmonic motion from one extreme position. If $a, b$ and $c$ are the displacements of the particle from the mean position at the ends of three successive seconds,the frequency of simple harmonic motion is

$A$ particle is performing simple harmonic motion along the $x$-axis with an amplitude of $4 \, cm$ and a time period of $1.2 \, s$. The minimum time taken by the particle to move from $x = 2 \, cm$ to $x = +4 \, cm$ and back again is given by .... $s$.

$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

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