(C) The displacement equation for $SHM$ is $y = A \sin (2t + \phi)$.The velocity is given by the derivative of displacement with respect to time: $v =

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(C) The displacement equation for $SHM$ is $y = A \sin (2t + \phi)$.The velocity is given by the derivative of displacement with respect to time: $v = \frac{dy}{dt} = 2A \cos (2t + \phi)$.At $t = 0$,$y = 2 \ m$ and $v = 4 \ ms^{-1}$.Substituting these values into the equations:$2 = A \sin (0 + \phi) \implies A \sin \phi = 2 \quad (i)$$4 = 2A \cos (0 + \phi) \implies A \cos \phi = 2 \quad (ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{A \sin \phi}{A \cos \phi} = \frac{2}{2}$$\tan \phi = 1$$\phi = \tan^{-1}(1) = 45^{\circ}$.

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

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The displacement of a particle executing simple harmonic motion is $y = A \sin (2t + \phi) \ m$,where $t$ is time in seconds and $\phi$ is the phase angle. At time $t = 0$,the displacement and velocity of the particle are $2 \ m$ and $4 \ ms^{-1}$ respectively. The phase angle $\phi$ is: (in $^{\circ}$)

AP EAMCET 2024 All AP EAMCET

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$60$
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$30$
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$45$
D
$90$

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Position of a Particle in SHM, Displacement and Phase

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DifficultKCET 2012

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The displacement of a simple harmonic oscillator after $3 \; s$ starting from its mean position is equal to half of its amplitude. The time period of the harmonic motion is $\dots \; s$.

MediumJEE MAIN 2022

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Medium

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The displacement-time equation of a particle executing $SHM$ is $x = A \sin \left( \omega t + \frac{\pi}{6} \right)$. The time taken by the particle to go directly from $x = -\frac{A}{2}$ to $x = +\frac{A}{2}$ is

Difficult

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$A$ simple harmonic oscillator has an amplitude $a$ and time period $T$. The time required by it to travel from $x = a$ to $x = a/2$ is

MediumAIPMT 1992

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