A particle is executing simple harmonic motio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The kinetic energy $K$ of a particle executing simple harmonic motion at a displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$,where $k$

Vedclass · Accepted Answer

$5: 4$

Vedclass · Answer

(C) The kinetic energy $K$ of a particle executing simple harmonic motion at a displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$,where $k$ is the force constant and $A$ is the amplitude.At displacement $x_1 = \frac{A}{4}$,the kinetic energy is $K_1 = \frac{1}{2} k (A^2 - (\frac{A}{4})^2) = \frac{1}{2} k (A^2 - \frac{A^2}{16}) = \frac{1}{2} k (\frac{15A^2}{16})$.At displacement $x_2 = \frac{A}{2}$,the kinetic energy is $K_2 = \frac{1}{2} k (A^2 - (\frac{A}{2})^2) = \frac{1}{2} k (A^2 - \frac{A^2}{4}) = \frac{1}{2} k (\frac{3A^2}{4}) = \frac{1}{2} k (\frac{12A^2}{16})$.The ratio of the kinetic energies is $\frac{K_1}{K_2} = \frac{\frac{15A^2}{16}}{\frac{12A^2}{16}} = \frac{15}{12} = \frac{5}{4}$.

$A$ particle is executing simple harmonic motion with amplitude $A$. The ratio of the kinetic energies of the particle when it is at displacements of $\frac{A}{4}$ and $\frac{A}{2}$ from the mean position is

Similar Questions

The mass of a particle is $1 \ kg$ and it is moving along the $x$-axis. The period of its oscillation is $\frac{\pi}{2} \ s$. Its potential energy at a displacement of $0.2 \ m$ is (in $J$)

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $6 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$

For any $S.H.M.$,the amplitude is $6\, cm$. If the instantaneous potential energy is half the total energy,then the distance of the particle from its mean position is .... $cm$.

The total energy of a particle performing $S.H.M.$ depends on:

Vedclass Test Series

Exam Paper Generator

Online Exam Module