A body is performing simple harmonic with an amplitude of $10 \,cm$. The velocity of the body was tripled by air Jet when it is at $5 \,cm$ from its mean position. The new amplitude of vibration is $\sqrt{ x } \,cm$. The value of $x$ is________
JEE MAIN 2022, Diffcult
Download our app for free and get started
$A =10 \,cm$
$\therefore$ Total Energy $=\frac{1}{2} KA ^{2}$
By energy conservation we can final $v$ at $x =5$
$\frac{1}{2} K (10)^{2}=\frac{1}{2} K (5)^{2}+\frac{1}{2} mv ^{2}$
$V =\sqrt{\frac{75 K }{ m }}$
Now, velocity is tripled through external mean so the amplitude of $SHM$ will charge and so the total energy, (but potential) energy at this moment will remain same)
$\therefore \frac{1}{2} K (5)^{2}+\frac{1}{2} m \left(3 \sqrt{\frac{75 K }{ m }}\right)^{2}=\frac{1}{2} KA ^{2}$
$\Rightarrow 25 K +675 K = KA ^{2}$
$\therefore A =\sqrt{700}$
$\therefore x =700$
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
- 1A particle executes $S.H.M.$ of amplitude A along $x$-axis. At $t =0$, the position of the particle is $x=\frac{A}{2}$ and it moves along positive $x$-axis the displacement of particle in time $t$ is $x=A \sin (\omega t+\delta)$, then the value $\delta$ will beView Solution
- 2For a particle showing motion under the force $F=-5(x-2)$, the motion is ...........View Solution
- 3Vertical displacement of a plank with a body of mass $'m'$ on it is varying according to law $y = \sin \omega t + \cos \omega t.$ The minimum value of $\omega $ for which the mass just breaks off the plank and the moment it occurs first after $t = 0$ are given by : ( $y$ is positive vertically upwards)View Solution
- 4View SolutionThe total energy of a particle executing S.H.M. is proportional to
- 5A mass $m$ is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are $k_1$ and $k_2$ respectively. When set into vertical vibrations, the period will beView Solution
- 6Two pendulums begins to swing simultaneously. The first pendulum makes $11$ full oscillations when the other makes $9$. The ratio of length of the two pendulums isView Solution
- 7A particle of mass $200 \,gm$ executes $S.H.M.$ The restoring force is provided by a spring of force constant $80 \,N / m$. The time period of oscillations is .... $\sec$View Solution
- 8View SolutionIf a watch with a wound spring is taken on to the moon, it
- 9Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and same mass $m$ is suspended with them. Now let $T$ be the time period in this position. ThenView Solution
- 10A 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