(C) The displacement function is $x(t) = A \sin (\omega t + \phi)$.At $t = 0$,$x(0) = A \sin \phi = 2 \dots (1)$.The velocity function is $v(t) = \fra

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(C) The displacement function is $x(t) = A \sin (\omega t + \phi)$.At $t = 0$,$x(0) = A \sin \phi = 2 \dots (1)$.The velocity function is $v(t) = \frac{dx}{dt} = A \omega \cos (\omega t + \phi)$.At $t = 0$,$v(0) = A \omega \cos \phi = 2 \omega \implies A \cos \phi = 2 \dots (2)$.Dividing equation $(1)$ by equation $(2)$:$\frac{A \sin \phi}{A \cos \phi} = \frac{2}{2} \implies \tan \phi = 1 \implies \phi = 45^{\circ}$.Substituting $\phi = 45^{\circ}$ in equation $(1)$:$A \sin 45^{\circ} = 2 \implies A \left( \frac{1}{\sqrt{2}} \right) = 2 \implies A = 2 \sqrt{2} \, cm$.Comparing $A = 2 \sqrt{2} \, cm$ with $x \sqrt{2} \, cm$,we get $x = 2$.

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

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$A$ particle executes simple harmonic motion represented by the displacement function $x(t) = A \sin (\omega t + \phi)$. If the position and velocity of the particle at $t = 0 \, s$ are $2 \, cm$ and $2 \omega \, cm \, s^{-1}$ respectively,then its amplitude is $x \sqrt{2} \, cm$,where the value of $x$ is ..... .

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A
$3$
B
$1$
C
$2$
D
$6$

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Oscillations

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Position of a Particle in SHM, Displacement and Phase

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$A$ particle performs $SHM$ about $x = 0$ such that at $t = 0$ it is at $x = 0$ and moving towards the positive extreme. The time taken by it to go from $x = 0$ to $x = \frac{A}{2}$ is ..... times the time taken to go from $x = \frac{A}{2}$ to $A$. The most suitable option for the blank space is

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$A$ particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards the left as shown in the figure. If the time period of $SHM$ executed about $O$ is $T$ and amplitude is $A$,then the equation of motion of the particle is:

Medium

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Consider a pair of identical pendulums,which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^o$ to the right with the vertical,the other pendulum makes an angle of $1^o$ to the left of the vertical. What is the phase difference between the pendulums?

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$A$ small mass executes linear $SHM$ about $O$ with amplitude $a$ and period $T.$ Its displacement from $O$ at time $T/8$ after passing through $O$ is:

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$A$ large number of random snapshots using a camera are taken of a particle in simple harmonic motion between $x = -x_0$ and $x = +x_0$,with the origin $x = 0$ as the mean position. $A$ histogram of the total number of times the particle is recorded about a given position (Event no.) would most closely resemble:

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$A$ particle performs $SHM$ about $x = 0$ such that at $t = 0$ it is at $x = 0$ and moving towards the positive extreme. The time taken by it to go from $x = 0$ to $x = \frac{A}{2}$ is ..... times the time taken to go from $x = \frac{A}{2}$ to $A$. The most suitable option for the blank space is

$A$ particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards the left as shown in the figure. If the time period of $SHM$ executed about $O$ is $T$ and amplitude is $A$,then the equation of motion of the particle is:

$A$ small mass executes linear $SHM$ about $O$ with amplitude $a$ and period $T.$ Its displacement from $O$ at time $T/8$ after passing through $O$ is:

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