When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{x}{8}$, where $x=$_____________.
JEE MAIN 2024, Diffcult
Download our app for free and get started
$ \text { Let total energy }=\mathrm{E}=\frac{1}{2} \mathrm{KA}^2 $
$ \mathrm{U}=\frac{1}{2} \mathrm{~K}\left(\frac{\mathrm{A}}{3}\right)^2=\frac{\mathrm{KA}^2}{2 \times 9}=\frac{\mathrm{E}}{9} $
$ \mathrm{KE}=\mathrm{E}-\frac{\mathrm{E}}{9}=\frac{8 \mathrm{E}}{9} $
$ \text { Ratio } \frac{\text { Total }}{\mathrm{KE}}=\frac{\mathrm{E}}{\frac{8 \mathrm{E}}{9}}=\frac{9}{8} $
$ \mathrm{x}=9$
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
- 1The total energy of the body executing $S.H.M.$ is $E$. Then the kinetic energy when the displacement is half of the amplitude, isView Solution
- 2Two equations of two $S.H.M.$ are $y = a\sin \,(\omega \,t - \alpha )$ and $y = b\cos (\omega \,t - \alpha )$. The phase difference between the two is .... $^o$View Solution
- 3A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, $x_1(t)=A \sin \omega t $ and $ x_2(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t)=B \sin (\omega t+\phi)$ brings the mass to a complete rest. The values of $B$ and $\phi$ areView Solution
- 4Two particles undergo $SHM$ along parallel lines with the same time period $(T)$ and equal amplitudes. At a particular instant, one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further timeView Solution
- 5Function $x$ = $A sin^2 wt + B cos^2 wt + C sin wt \ cos wt$ does not represents $SHM$ for this conditionView Solution
- 6The displacement time equation of a particle executing $SHM$ is : $x = A \,sin\,(\omega t + \phi )$. At time $t = 0$ position of the particle is $x = A/2$ and it is moving along negative $x-$ direction. Then the angle $\phi $ can beView Solution
- 7A 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
- 8The position co-ordinates of a particle moving in a $3-D$ coordinates system is given by $x = a\,cos\,\omega t$ , $y = a\,sin\,\omega t$ and $z = a\omega t$ The speed of the particle isView Solution
- 9What is the maximum acceleration of the particle doing the $SHM$ $y = 2\sin \left[ {\frac{{\pi t}}{2} + \phi } \right]$ where $2$ is in $cm$View Solution
- 10The displacement of a particle is given at time $t$, by:$x=A \sin (-2 \omega t)+B \sin ^2 \omega t \quad$ Then,View Solution