A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is on
JEE MAIN 2023, Medium
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 of mass $10\, gm$ moves in a field where potential energy per unit mas is given by expression $v = 8 \times 10^4\, x^2\, erg/gm$. If the total energy of the particle is $8 \times 10^7\, erg$ then the relation between $x$ and time $t$ isView Solution
- 2The length of simple pendulum is increased by $44\%$. The percentage increase in its time period will be ..... $\%$View 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
- 4If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, areView Solution
- 5A particle performing $SHM$ is found at its equilibrium at $t = 1\,sec$. and it is found to have a speed of $0.25\,m/s$ at $t = 2\,sec$. If the period of oscillation is $6\,sec$. Calculate amplitude of oscillationView Solution
- 6A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to equation $y = k{t^2}$ where $k = 1\,m/se{c^2}$. If new time period is $T_2$ then ratio $\frac{{T_1^2}}{{T_2^2}}$ will beView Solution
- 7A simple pendulum with iron bob has a time period $T$. The bob is now immersed in a non-viscous liquid and oscillated. If the density of liquid is $\frac{1}{12}$ th that of iron, then new time period will beView Solution
- 8The potential energy of a particle of mass $100 \,g$ moving along $x$-axis is given by $U=5 x(x-4)$, where $x$ is in metre. The period of oscillation is .................View Solution
- 9The period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\alpha$, is given byView Solution
- 10The displacement of a particle executing simple harmonic motion is given byView Solution
$\mathrm{y}=\mathrm{A}_{0}+\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}$
Then the amplitude of its oscillation is given by