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Energy in SHM is considered one the most difficult concept.
50 Questions around this concept.
For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?
(graphs are schematic and not drawn to scale)
In a simple harmonic oscillator, at the mean position
A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement . Which of the following statements is true?
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A particle of mass executes simple harmonic motion with amplitude
and frequency
. The average kinetic energy during its motion from the position of equilibrium to the end is :
A particle is executing simple harmonic motion with a time period T. At time t=0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like :
The total energy of a particle, executing simple harmonic motion is:
The kinetic energy in an SHM is best represented by.
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A particle of mass m is executing SHM about its mean position. The total energy of the particle at a given instant is:
A particle oscillates along the x -axis according to the law, $x(t)=x_0 \sin ^2\left(\frac{t}{2}\right)$ where $x_0=1 \mathrm{~m}$. The kinetic energy $(\mathrm{K})$ of the particle as a function of $x$ is correctly represented by the graph.
A particle executing S.H.M. possesses two types of energy: Potential energy and Kinetic energy
Potential energy-
As restoring force is given as $F=-k x$
$
\begin{aligned}
U & =-\int d w=-\int_0^x F d x=\int_0^x k x d x=\frac{1}{2} k x^2 \\
\text { using } \omega & =\sqrt{\frac{k}{m}} \text { or } k=m \omega^2
\end{aligned}
$
we get $U=\frac{1}{2} m \omega^2 x^2$
For $x=A \sin (w t)$
$
U=\frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t
$
- Potential energy maximum and equal to total energy at extreme positions
$
U_{\max }=\frac{1}{2} k A^2=\frac{1}{2} m \omega^2 A^2 \quad \text { when } x= \pm A ; \omega t=\pi / 2 ; \quad t=T / 4
$
- Potential energy is minimum at mean position
i.e $U_{\min }=0 \quad$ when $x=0 ; \omega t=0 ; t=0$
- The average value of potential energy with respect to $t$
$
\begin{aligned}
& \text { Average of } U=\frac{\int U d t}{\int d t} \\
& \because U=\frac{1}{2} k x^2 \\
& U_{\text {avg }}=\frac{\int \frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t}{\int d t}=\frac{\int \frac{1}{4} m \omega^2 A^2(1-\cos 2 \omega t) d t}{d t}=\frac{1}{4} m \omega^2 A^2
\end{aligned}
$
Kinetic energy-
$
K=\frac{1}{2} m v^2
$
or using $v=A \omega \cos \omega t$ we get $K=\frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t$
And using $v=w \sqrt{A^2-x^2}$ and $k=m \omega^2$ we get
$
K . E .=\frac{1}{2} K\left(A^2-x^2\right)
$
- Kinetic energy is maximum at the mean position and equal to total energy at the mean position.
$
\text { i.e } K_{\max }=\frac{1}{2} m \omega^2 A^2 \quad \text { when } x=0 ; t=0 ; \omega t=0
$
- Kinetic energy is minimum at the extreme positions.
i.e $K_{\min }=0 \quad$ when $y=A ; t=T / 4, \omega t=\pi / 2$
- The average value of kinetic energy with respect to $t$
$
\begin{gathered}
K_{\text {avg }}=\frac{\int K d t}{\int d t} \\
K_{\text {avg }}=\frac{\int \frac{1}{2} m \omega^2 A^2 \cos ^2(\omega t)}{\int d t}=\frac{\int \frac{1}{4} m \omega^2 A^2(1+\cos 2 \omega t) d t}{d t}=\frac{1}{4} m \omega^2 A^2
\end{gathered}
$
So $K_{\text {avg }}=U_{\text {avg }}$
Total energy-
- Total mechanical energy $=$ Kinetic energy + Potential energy or $\mathrm{E}=\mathrm{K}+\mathrm{U}$
$
E=\frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \omega^2 A^2
$
So Total energy does not depend on position(x) i.e. it always remains constant in SHM.
At time t=0 sec, the position of the block is equal to the amplitude,
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