A simple harmonic oscillator has an amplitude $A$ and time period $6 \pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x}$ s, where $x=$__________.
JEE MAIN 2024, Diffcult
Download our app for free and get started
From phasor diagram particle has to move from $\mathrm{P}$ to $Q$ in a circle of radius equal to amplitude of $SHM.$
$ \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} $
$ \phi=\frac{\pi}{6}$
Now, $\frac{\pi}{6}=\omega t$
$\frac{\pi}{6}=\frac{2 \pi}{T} t$
$\frac{\pi}{6}=\frac{2 \pi}{6 \pi} t$
$\mathrm{t}=\frac{\pi}{2}$
So, $\mathrm{x}=2$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1Two simple harmonic motions $y_1 = A \sin \omega t$ and $y_2 =A \cos \omega t$ are superimposed on a particle of mass $m.$ The total mechanical energy of the particle is :View Solution
- 2The kinetic energy of a particle executing $S.H.M.$ is $16\, J$ when it is in its mean position. If the amplitude of oscillations is $25\, cm$ and the mass of the particle is $5.12\, kg$, the time period of its oscillation isView Solution
- 3A particle moves in the $x-y$ plane according to equation $\overrightarrow r = (\widehat i + 2\widehat j)\, A \, \cos \omega t$. The motion of the particle isView Solution
- 4A block of mass $m$ hangs from three springs having same spring constant $k$. If the mass is slightly displaced downwards, the time period of oscillation will beView Solution
- 5The $x$-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t=4 / 3 \mathrm{~s}$ isView Solution
- 6A cylindrical block of wood (density $= 650\, kg\, m^{-3}$), of base area $30\,cm^2$ and height $54\, cm$, floats in a liquid of density $900\, kg\, m^{-3}$ . The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length ..... $cm$ (nearly)View Solution
- 7A body performs $S.H.M.$ Its kinetic energy $K$ varies with time $t$ as indicated by graphView Solution
- 8A vibratory motion is represented by $x = 2A\,\cos \omega t + A\,\cos \,\left( {\omega t + \frac{\pi }{2}} \right) + A\,\cos \,\left( {\omega t + \pi } \right)$ $ + \frac{A}{2}\,\cos \left( {\omega t + \frac{{3\pi }}{2}} \right)$. The resultant amplitude of the motion isView Solution
- 9A particle executes $S.H.M.$ and its position varies with time as $x=A$ sin $\omega t$. Its average speed during its motion from mean position to mid-point of mean and extreme position isView Solution
- 10A bob of mass $'m'$ suspended by a thread of length $l$ undergoes simple harmonic oscillations with time period ${T}$. If the bob is immersed in a liquid that has density $\frac{1}{4}$ times that of the bob and the length of the thread is increased by $1 / 3^{\text {rd }}$ of the original length, then the time period of the simple harmonic oscillations will be :-View Solution