A particle starts oscillating simple harmonic | vedclass

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

Question

(D) The displacement of a particle starting from the mean position is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega = 2\pi/T$

Vedclass · Accepted Answer

$3: 1$

Vedclass · Answer

(D) The displacement of a particle starting from the mean position is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega = 2\pi/T$ is the angular frequency.At $t = T/6$,the displacement is $x = A \sin((2\pi/T) \cdot (T/6)) = A \sin(\pi/3) = A \sqrt{3}/2$.The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k (A \sqrt{3}/2)^2 = \frac{3}{8} k A^2$.The total energy $E$ is given by $E = \frac{1}{2} k A^2$.The kinetic energy $K$ is $K = E - U = \frac{1}{2} k A^2 - \frac{3}{8} k A^2 = \frac{1}{8} k A^2$.The ratio of potential energy to kinetic energy is $U/K = (3/8 k A^2) / (1/8 k A^2) = 3/1$.Thus,the ratio is $3: 1$.

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$

Similar Questions

$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x+y)$ is

$A$ body is executing $S.H.M$. At a displacement $x$ its potential energy is $9 \ J$ and at a displacement $y$ its potential energy is $16 \ J$. The potential energy at displacement $(x+y)$ is (in $J$)

The displacements of two particles of same mass executing $SHM$ are represented by the equations $x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$ and $x_2=5 \cos (\omega t)$. The value of $\omega$ for which the energies of both the particles remain same is (in $\text{ unit}$)

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal,when displacement (amplitude = $a$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module