The total energy of a body executing S.H.M. i | vedclass

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

Question

(C) The total energy of a body executing $S.H.M.$ is given by $E = \frac{1}{2} k a^2$,where $a$ is the amplitude.The potential energy at a displacemen

Vedclass · Accepted Answer

$\frac{3E}{4}$

Vedclass · Answer

(C) The total energy of a body executing $S.H.M.$ is given by $E = \frac{1}{2} k a^2$,where $a$ is the amplitude.The potential energy at a displacement $y$ is given by $U = \frac{1}{2} k y^2$.The kinetic energy $K$ is given by $K = E - U = \frac{1}{2} k a^2 - \frac{1}{2} k y^2 = \frac{1}{2} k (a^2 - y^2)$.Given that the displacement $y = \frac{a}{2}$,we substitute this into the kinetic energy formula:$K = \frac{1}{2} k (a^2 - (\frac{a}{2})^2)$$K = \frac{1}{2} k (a^2 - \frac{a^2}{4})$$K = \frac{1}{2} k (\frac{3a^2}{4})$$K = \frac{3}{4} (\frac{1}{2} k a^2)$Since $E = \frac{1}{2} k a^2$,we get $K = \frac{3E}{4}$.

The total energy of a body executing $S.H.M.$ is $E$. Then the kinetic energy when the displacement is half of the amplitude,is

Similar Questions

The total energy of a particle executing $S.H.M.$ is proportional to

For the minimum time period condition,find the average potential energy between $t = 0$ to $t = 0.05 \ s$ (take $g = 10 \ m/s^2$).

$A$ body is executing a linear $S.H.M.$ Its potential energies at the displacements $x$ and $y$ are $E_1$ and $E_2$ respectively. Its potential energy at displacement $(x+y)$ will be

The maximum potential energy of a block executing simple harmonic motion is $25 \ J$. $A$ is the amplitude of oscillation. At $x = A / 2$,the kinetic energy of the block is $...............$ (in $J$)

For a body performing simple harmonic motion,its potential energy is $E_x$ at displacement $x$ and $E_y$ at displacement $y$ from the mean position. The potential energy $E_0$ at displacement $(x+y)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module