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Equations of motions of SHM is considered one the most difficult concept.
Simple harmonic motion is considered one of the most asked concept.
94 Questions around this concept.
The restoring force of SHM is maximum when the particle is:
A particle is moving with constant angular velocity along the circumference of a circle. Which of the following statement is true
The motion which is not simple harmonic is
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A simple harmonic motion is represented by: $ y=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) \mathrm{cm} $
The amplitude and time period of the motion are:
The displacement of an SHM is given by
$y=3 \operatorname{Sin}(w t+\phi)$
The acceleration is
The equation of SHM is $y=2 \sin (2 \pi r t+\phi)$ then its phase at time t is.
The locus of resultant of the SHM $y=3 \sin \omega t \quad x=2 \sin \omega t$
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The locus of the resultant of $x=3 \sin \omega t, y=2 \sin \left(\omega t+\frac{\pi}{2}\right)$ is
The position at which amplitude of oscillation has zero value are called
The ratio of the total energy of SHM with amplitude A and time period T at displacement equals half the amplitude and one-fourth of the amplitude is
Periodic motion is also called as harmonic motion.
Simple harmonic motion is the simplest form of oscillatory motion in which the particle oscillates on a straight line and the restoring force is always directed towards the mean position and it’s magnitude at any instant is directly proportional to the displacement of the particle from the mean position at that instant i.e. Restoring force α Displacement of the particle from mean position.
F = -Kx, where x is measured from mean position
All oscillations are not simple harmonic motions but all simple harmonic motions are oscillatory motions.
Let’s understand SHM with the help of spring block system:
Suppose we stretch the spring to the extreme position and then release it from there.
Here we can see that acceleration is always directed towards the mean position.
And $F=-k x$
Also, $a=\frac{F}{m} \Rightarrow a=-\frac{k}{m} x \Rightarrow a=-\omega^2 x ;$ where $\omega^2$ is a positive constant and $\omega=\sqrt{\frac{k}{m}} \Rightarrow k=m \omega^2$,
where k is force or spring constant.
- $\mathrm{v}=0$ at extreme position
- $\mathrm{v}=\mathrm{max}$ at mean position
- $a=0$ at mean position
- $a=\max$ at extreme position, i.e., at $x= \pm A, a= \pm \omega^2 A$
- Magnitude of maximum acceleration, $\left|a_{\max }\right|=\omega^2 A$
As we know, $a=-\omega^2 x$
$
\Rightarrow \frac{d v}{d t}=-\omega^2 x \Rightarrow v \frac{d v}{d x}=-\omega^2 x \Rightarrow v d v=-\omega^2 x d x
$
Let the particle is released from an extreme position, i.e., at $x=+A, v=0$ and it becomes $v$ when the displacement becomes x .
On integrating both sides of the above equation, we get:
$
\begin{aligned}
& \int_0^v v d v=\int_A^x-\omega^2 x d x \\
& \Rightarrow\left[\frac{v^2}{2}\right]_0^v=-\omega^2\left[\frac{x^2}{2}\right]_A^x \\
& \Rightarrow v^2-0=-\omega^2\left(x^2-A^2\right) \\
& \Rightarrow v^2=\omega^2\left(A^2-x^2\right) \\
& \Rightarrow v= \pm \omega \sqrt{\left(A^2-x^2\right)} \\
& \text { At } x=0, v_{\max }= \pm \omega A
\end{aligned}
$
Note-
As the relation between velocity ( v ) and position ( x ) in SHM is given by
$
v= \pm \omega \sqrt{\left(A^2-x^2\right)}
$
This can be rearranged as
$\begin{aligned} & v^2=\omega^2\left(A^2-x^2\right) \\ \Rightarrow & v^2=\omega^2 A^2-\omega^2 x^2 \\ \Rightarrow & v^2+\omega^2 x^2=\omega^2 A^2 \\ \Rightarrow & \frac{v^2}{\omega^2 A^2}+\frac{x^2}{A^2}=1\end{aligned}$
And this shows that the velocity-position graph is an ellipse (as shown in the below figure)
where
Major axis=2A
and Minor axis $=2 \omega A$
- General equation of SHM
1. For Displacement:-
$x=A S i n(w t+\phi) ;$ where $\phi$ is initial phase or epoch and $(\omega t+\phi)$ is called as phase.
Various displacement equations:-
(1) $x=$ ASinwt $\Rightarrow$ when particle starts from mean position towards right.
(2) $x=-$ ASinwt $\Rightarrow$ when particle starts from mean position towards left.
(3) $x=A \operatorname{Coswt} \Rightarrow$ when particle starts from right extreme position towards left
(4) $x=-$ ACoswt $\Rightarrow$ when particle starts from left extreme position towards Right
2. For Velocity (v):-
$
\begin{aligned}
x & =A \operatorname{Sin}(\omega t+\phi) \\
\Rightarrow v & =\frac{d x}{d t}=A \omega \operatorname{Cos}(\omega t+\phi)=A \omega \operatorname{Sin}\left(\omega t+\phi+\frac{\pi}{2}\right)
\end{aligned}
$
3. For Acceleration:-
$
\begin{aligned}
x & =A \operatorname{Sin}(\omega t+\phi) \\
\Rightarrow v & =\frac{d x}{d t}=A \omega \operatorname{Cos}(\omega t+\phi)=A \omega \operatorname{Sin}\left(\omega t+\phi+\frac{\pi}{2}\right) \\
\Rightarrow a & =\frac{d v}{d t}=-A \omega^2 \operatorname{Sin}(\omega t+\phi)=A \omega^2 \operatorname{Sin}(\omega t+\phi+\pi)=-\omega^2 x
\end{aligned}
$
So here we can see that the phase difference between x and v is $\frac{\pi}{2}$ similarly, the phase difference between v and a is $\frac{\pi}{2}$
similarly, the phase difference between a and x is
Differential equation of SHM:-
$
\begin{aligned}
& \frac{d v}{d t}=-\omega^2 x \\
& \Rightarrow \frac{d}{d t}\left(\frac{d x}{d t}\right)=-\omega^2 x \\
& \Rightarrow \frac{d^2 x}{d t^2}+\omega^2 x=0
\end{aligned}
$
If the motion of any particle satisfies this equation then that particle will do SHM.
- Different graphs in SHM
For $x=a \operatorname{Sin}(\omega t)$
Graph of displacement v/s time is given as
Graph of velocity V/s time
$v=a \omega \operatorname{Cos}(\omega t)$
Graph of acceleration V/s time
acceleration $=-a \omega^2 \operatorname{Sin}(\omega t)$
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