The maximum potential energy of a block executing simple harmonic motion is $25\,J$. A is amplitude of oscillation. At $A / 2$, the kinetic energy of the block is $...............$
JEE MAIN 2023, Easy
Download our app for free and get started
$u _{\max }=\frac{1}{2} m \omega^2 A ^2=25\,J$
$KE \text { at } \frac{ A }{2}=\frac{1}{2} mv _1^2=\frac{1}{2} m \omega^2\left( A ^2-\frac{ A ^2}{4}\right)$
$KE =\frac{1}{2} m \omega^2 \frac{3 A ^2}{4}=\frac{3}{4}\left(\frac{1}{2} m \omega^2 A ^2\right)$
$KE =\frac{3}{4} \times 25=18.75\,J$
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
- 1An object of mass $0.2 \mathrm{~kg}$ executes simple harmonic motion along $\mathrm{x}$ axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $\mathrm{x}=0.04 \mathrm{~m}$ the object has kinetic energy $0.5 \mathrm{~J}$ and potential energy $0.4 \mathrm{~J}$ The amplitude of oscillation is ............ cm.View Solution
- 2The motion of a simple pendulum excuting $S.H.M$. is represented by following equation.View Solution
$Y = A \sin (\pi t +\phi)$, where time is measured in $second$.
The length of pendulum is .............$cm$
- 3A particle executes simple harmonic motion between $x =- A$ and $x =+ A$. If time taken by particle to go from $x=0$ to $\frac{A}{2}$ is $2 s$; then time taken by particle in going from $x =\frac{ A }{2}$ to $A$ is $.........\,s$View Solution
- 4View SolutionIf a particle is executing simple harmonic motion, then acceleration of particle
- 5A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$, then the time period is given by $T = 2\pi \sqrt {\frac{l}{{g'}}} $, where $g'$ is equal toView Solution
- 6The piston in the cylinder head of locomotive has a stroke of $6\,m$ (which is twice the amplitude). If the piston executing simple harmonic motion with an angular frequency of $200\, rad\, min^{-1}$, its maximum speed is .... $ms^{-1}$View Solution
- 7A 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
- 8A particle in $SHM $ is described by the displacement equation $x(t) = A\cos (\omega t + \theta ).$ If the initial $(t = 0)$ position of the particle is $1 \,cm$ and its initial velocity is $\pi $cm/s, what is its amplitude? The angular frequency of the particle is $\pi {s^{ - 1}}$View Solution
- 9A particle executes simple harmonic motion with an amplitude of $4 \,cm$. At the mean position the velocity of the particle is $10\, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \,cm/s$ isView Solution
- 10The acceleration $a$ of a particle undergoing $S.H.M.$ is shown in the figure. Which of the labelled points corresponds to the particle being at -$x_{max}$View Solution
