Simple Harmonic Motion Concept Page - 4

Result

Maximum and Minimum acceleration in SHM

General equation of acceleration is:
a=−Aw2sinwt$a=-A{w}^{2}sinwt$

From this expression one could infer directly that acceleration is maximum for:
wt=90$wt=90$ ( at the extreme position)
Acceleration is minimum for:
wt=0$wt=0$ ( at the mean position)

Diagram

Draw a plot of acceleration as a function of time

Above graph clearly depicts variation of acceleration with time in a simple harmonic motion.

Diagram

Draw a plot of acceleration as a function of distance

Since, a=−w2x$a=-w^{2}x$
a$a$ Vs x$x$ plot essentially represents a straight line with slope −w2$-w^2$ . 

Definition

Angular Acceleration as a function of time

α=−w2θ$\alpha =-w^2\theta$
Angular acceleration as a function of time in SHM:
α=−w2θosinwt$\alpha =-w^2{\theta }_{o}sinwt$
Since,
 θ=θosinwt$\theta ={\theta }_{o}sinwt$

Diagram

Plot of displacement as a function of time and interpret amplitude, phase and time period

From the above graph we can infer that:
Peak of the curve goes to height A, hence it's amplitude is A.
Time period is T, as it takes time T to complete one cycle.
Phase for this SHM is zero.

Formula

Differential Equation of SHM

d2xdt=−w2x${\displaystyle \frac{d^{2}x}{dt}}=-w^{2}x$
General solution to this equation is:

x=Asin(wt)+Bcos(wt)$x=Asin(wt)+Bcos(wt)$
On Putting the boundary conditions specific to the given problem we get:
x=Asin(wt+δ)$x=Asin(wt+\delta )$

Example

Find time period or frequency by finding equation of SHM using differentiation of energy equation

Example: A particle of mass m$m$ is moving in a field where the potential energy is given by U(x)=U0(1−cosax)$U(x)=U_0(1-\cos ax)$, where U0$U_0$ and a$a$ are positive constants and x$x$ is the displacement from mean position. Then (for small oscillations) find the time period of oscillation?

Solution:
U(x)=U0(1−cosax)$U(x)=U_0(1-\cos ax)$
dUdx=U0asinax${\displaystyle \frac{dU}{dx}}=U_{0}a\sin ax$
F=−dUdx;F=−U0asinax$F=-{\displaystyle \frac{dU}{dx}};F=-U_{0}a\sin ax$
For small x$x$          F=−U0a2x$F=-U_0{a}^{2}x$
       acceleration=−U0a2mx;T=2πmU0a2−−−−−√$acceleration=-{\displaystyle \frac{U_0{a}^2}{m}}x;T=2\pi \sqrt{{\displaystyle \frac{m}{U_0{a}^2}}}$
At x=0,F=0$x=0,F=0$ hence mean position and speed of particle is maximum.

Formula

Differential equation of linear wave

The linear wave equation in three dimensions:
δ2qδt2=c2(δ2qδx2+δ2qδy2+δ2qδz2)${\displaystyle \frac{{\delta }^{2}q}{\delta t^2}}=c^2({\displaystyle \frac{{\delta }^{2}q}{\delta x^2}}+{\displaystyle \frac{{\delta }^{2}q}{\delta y^2}}+{\displaystyle \frac{{\delta }^{2}q}{\delta z^2}})$
where, q=q(x,y,z,t)$q=q(x,y,z,t)$ and x,y,z$x,y,z$ are standard Cartesian coordinates.

Law

State and use the condition on force for simple harmonic motion

There is no loss of energy in the simple harmonic motion. In order to sustain this constraint the applied force must be conservative.
For example, Spring force is conservative and friction is non-conservative.

Formula

Use the relation between restoring force and potential energy

Restoring force is given by: F=−dUdx$F=-{\displaystyle \frac{dU}{dx}}$
It is often useful to find the equation of SHM.
Example:
A particle of mass 10$10$ gm is placed in a potential field given by V=(50x2+100)J/kg$V=(50x^2+100)J/kg$. Find the frequency of oscillation in cycle/sec.Solution:
Potential energy U=mV$U=mV$
⇒U=(50x2+100)10−2$\Rightarrow U=(50x^2+100){10}^{-2}$
F=−dUdx=−(100x)10−2$F=-{\displaystyle \frac{dU}{dx}}=-(100x){10}^{-2}$
⇒mω2x=−(100×10−2)x$\Rightarrow m{\omega }^{2}x=-(100\times {10}^{-2})x$
10×10−3ω2x=100×10−2x$10\times {10}^{-3}{\omega }^{2}x=100\times {10}^{-2}x$
⇒ω2=100,ω=10$\Rightarrow {\omega }^2=100,\omega =10$
⇒f=ω2π=102π=5π$\Rightarrow f={\displaystyle \frac{\omega }{2\pi }}={\displaystyle \frac{10}{2\pi }}={\displaystyle \frac{5}{\pi }}$

Definition

Relation between restoring torque and potential energy

Restoring torque of a SHM can be found by: τ=−dUdθ$\tau =-{\displaystyle \frac{dU}{d\theta }}$
It is often useful in finding equation of SHM and helps in solving problems.

Formula

Potential energy per unit length at a given point in a travellling sound wave

Potential energy per unit length  (w)$(w)$ at a point is defined as:

w=pvc$w={\displaystyle \frac{pv}{c}}$
p is the sound pressure.
v is the particle velocity in the direction of propagation.
c is the speed of sound.

BookMarks

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