(B) The displacement of a particle in simple harmonic motion starting from the equilibrium position is given by $x(t) = a \sin(\omega t)$.We need to f

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(B) The displacement of a particle in simple harmonic motion starting from the equilibrium position is given by $x(t) = a \sin(\omega t)$.We need to find the time $t$ when the displacement $x(t) = a/2$.Substituting the values,we get: $a/2 = a \sin(\omega t)$.This simplifies to $\sin(\omega t) = 1/2$.Since $\sin(\pi/6) = 1/2$,we have $\omega t = \pi/6$.Substituting $\omega = 2\pi/T$,we get $(2\pi/T) \cdot t = \pi/6$.Solving for $t$,we find $t = T/12$.

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

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$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

AIPMT 2007 All AIPMT

A
$T/8$
B
$T/12$
C
$T/2$
D
$T/4$

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Oscillations

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Position of a Particle in SHM, Displacement and Phase

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$A$ particle is executing simple harmonic motion with a period of $T$ seconds and amplitude $a$ meters. The shortest time it takes to reach a point $\frac{a}{\sqrt{2}} \, m$ from its mean position in seconds is

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

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The amplitude and the periodic time of a $S.H.M.$ are $5 \, cm$ and $6 \, s$ respectively. At a distance of $2.5 \, cm$ away from the mean position,the phase will be

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$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

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$A$ particle is executing simple harmonic motion with a period of $T$ seconds and amplitude $a$ meters. The shortest time it takes to reach a point $\frac{a}{\sqrt{2}} \, m$ from its mean position in seconds is

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