(B) Given the displacement equation $x(t) = x_m \cos(\omega t + \phi)$.At $t = 0$,$x = -x_m$,so $-x_m = x_m \cos(\phi)$,which implies $\cos(\phi) = -1

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(B) Given the displacement equation $x(t) = x_m \cos(\omega t + \phi)$.At $t = 0$,$x = -x_m$,so $-x_m = x_m \cos(\phi)$,which implies $\cos(\phi) = -1$,so $\phi = \pi$.The equation becomes $x(t) = x_m \cos(\omega t + \pi) = -x_m \cos(\omega t)$.We want to find $t$ when $x = +x_m$.So,$x_m = -x_m \cos(\omega t)$,which means $\cos(\omega t) = -1$.This occurs when $\omega t = \pi$ (for the first time).Since $\omega = \frac{2\pi}{T}$,we have $\frac{2\pi}{T} \cdot t = \pi$.Solving for $t$,we get $t = \frac{T}{2} = 0.50\, T$.

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

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The displacement of a particle undergoing $SHM$ with time period $T$ is given by $x(t) = x_m \cos(\omega t + \phi)$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = +x_m$ when:

AIIMS 2011 All AIIMS

A
$t = 0.25\, T$
B
$t = 0.50\, T$
C
$t = 0.75\, T$
D
$t = 1.00\, T$

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Oscillations

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Position of a Particle in SHM, Displacement and Phase

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The time period of a particle executing $S.H.M.$ is $8 \,s$. At $t=0$ it is at the mean position. The ratio of distance covered by the particle in the $1^{\text{st}}$ second to the $2^{\text{nd}}$ second is:

Medium

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$A$ particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.

MediumMHT CET 2011

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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 displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.

Medium

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What is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t)$?

EasyWBJEE 2012

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The displacement of a particle undergoing $SHM$ with time period $T$ is given by $x(t) = x_m \cos(\omega t + \phi)$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = +x_m$ when:

Similar Questions

The time period of a particle executing $S.H.M.$ is $8 \,s$. At $t=0$ it is at the mean position. The ratio of distance covered by the particle in the $1^{\text{st}}$ second to the $2^{\text{nd}}$ second is:

$A$ particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.

$A$ 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 displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.

What is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t)$?

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