(C) The phase of a particle in $SHM$ is defined by the argument of the sine or cosine function in the displacement equation,$x(t) = A \sin(\omega t +

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Both the position and direction of motion of the particle at time $t$

(C) The phase of a particle in $SHM$ is defined by the argument of the sine or cosine function in the displacement equation,$x(t) = A \sin(\omega t + \phi)$.This phase angle $(\omega t + \phi)$ uniquely determines the state of the particle.Specifically,it provides information about both the displacement (position) of the particle relative to the equilibrium position and the direction of its velocity at that specific instant $t$.

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

Easy

The phase (at a time $t$) of a particle in simple harmonic motion tells:

A
Only the position of the particle at time $t$
B
Only the direction of motion of the particle at time $t$
C
Both the position and direction of motion of the particle at time $t$
D
Neither the position of the particle nor its direction of motion at time $t$

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Oscillations

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Position of a Particle in SHM, Displacement and Phase

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If the time taken by the particle to go from $x = 0$ to $x = A/2$ is $2 \, s$,then the time taken by the particle in going from $x = A/2$ to $x = A$ is $......... \, s$.

MediumJEE MAIN 2023

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$A$ particle executes simple harmonic motion (amplitude $= A$) between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is ${T_1}$ and to go from $x = A/2$ to $x = A$ is ${T_2}$. Then:

MediumIIT 2001

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The periodic time of a particle doing simple harmonic motion is $4 \,s$. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

EasyMHT CET 2009

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What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

Easy

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The equation of $S.H.M.$ is $y = a \sin(2\pi nt + \alpha)$,then its phase at time $t$ is:

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The phase (at a time $t$) of a particle in simple harmonic motion tells:

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$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If the time taken by the particle to go from $x = 0$ to $x = A/2$ is $2 \, s$,then the time taken by the particle in going from $x = A/2$ to $x = A$ is $......... \, s$.

$A$ particle executes simple harmonic motion (amplitude $= A$) between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is ${T_1}$ and to go from $x = A/2$ to $x = A$ is ${T_2}$. Then:

The periodic time of a particle doing simple harmonic motion is $4 \,s$. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

The equation of $S.H.M.$ is $y = a \sin(2\pi nt + \alpha)$,then its phase at time $t$ is:

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