(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

Difficult

$A$ 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

$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

MediumMHT CET 2025

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$A$ particle initially at the mean position is executing simple harmonic motion with an angular frequency $\omega = \frac{\pi}{4} \text{ rad s}^{-1}$. The ratio of the distances travelled by the particle in the first second and the second second is:

DifficultAP EAMCET 2023

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$A$ particle is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

Medium

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Two particles are executing $S.H.M.$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions,each time their displacement is half of their amplitude. What is the phase difference between them?

Difficult

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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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$A$ 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

$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

$A$ particle initially at the mean position is executing simple harmonic motion with an angular frequency $\omega = \frac{\pi}{4} \text{ rad s}^{-1}$. The ratio of the distances travelled by the particle in the first second and the second second is:

$A$ particle is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

Two particles are executing $S.H.M.$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions,each time their displacement is half of their amplitude. What is the phase difference between them?

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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