(B) The displacement function is given by $x(t) = a \cos(\omega t + \theta)$.At $t = 0$,$x(0) = a \cos \theta = 1 \, cm$ ---$(i)$The velocity functio

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(B) The displacement function is given by $x(t) = a \cos(\omega t + \theta)$.At $t = 0$,$x(0) = a \cos \theta = 1 \, cm$ ---$(i)$The velocity function is $v(t) = \frac{dx}{dt} = -a\omega \sin(\omega t + \theta)$.At $t = 0$,$v(0) = -a\omega \sin \theta = \pi \, cm/s$.Given $\omega = \pi \, rad/s$,we have $-a\pi \sin \theta = \pi$,which simplifies to $a \sin \theta = -1 \, cm$ ---(ii)Squaring and adding equations $(i)$ and (ii):$(a \cos \theta)^2 + (a \sin \theta)^2 = (1)^2 + (-1)^2$$a^2(\cos^2 \theta + \sin^2 \theta) = 1 + 1$$a^2 = 2$$a = \sqrt{2} \, cm$.

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

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$A$ particle in $S.H.M.$ is described by the displacement function $x(t) = a\cos (\omega t + \theta )$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$. The angular frequency of the particle is $\pi \, rad/s$,then its amplitude is

A
$1 \, cm$
B
$\sqrt{2} \, cm$
C
$2 \, cm$
D
$2.5 \, cm$

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Oscillations

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Position of a Particle in SHM, Displacement and Phase

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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Two particles $A$ and $B$ execute $SHMs$ of periods $T$ and $\frac{3T}{2}$. If they start from the mean position,then the phase difference between them,when the particle $A$ completes two oscillations,will be

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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.

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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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$A$ particle executes simple harmonic motion. Its amplitude is $8 \,cm$ and time period is $6 \,s$. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude,is ............. $s$

MediumJEE MAIN 2022

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$A$ particle in $S.H.M.$ is described by the displacement function $x(t) = a\cos (\omega t + \theta )$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$. The angular frequency of the particle is $\pi \, rad/s$,then its amplitude is

Similar Questions

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

Two particles $A$ and $B$ execute $SHMs$ of periods $T$ and $\frac{3T}{2}$. If they start from the mean position,then the phase difference between them,when the particle $A$ completes two oscillations,will be

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.

$A$ 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

$A$ particle executes simple harmonic motion. Its amplitude is $8 \,cm$ and time period is $6 \,s$. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude,is ............. $s$

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