Simple Harmonic Motion (SHM)

Simple Harmonic Motion

It is the periodic (repeated) motion. in which a particle moves to and fro motion about a mean (equilibrium) position under a restoring force(acceleration) which is always directed toward mean position such that its magnitude is proportional to displacement (a∝ -y).

Or, F ∝ -y

$\therefore$F= -ky

 Types: (i) Linear SHM; a∝-y

           Or, a=-ky

           Or,$\frac{d^{2}y}{d{t}^2}+Ky=0$

(ii) Angular SHM

∝ = -Kθ

            $\frac{d^2\theta }{d{t}^2}+K\theta =0$                

Equation of Simple Harmonic Motion in different conditions

  The general equation of SHM is α= Asin(wt+$\varphi$).

1) From position (x=0, t=0) to x=+A

X= Asinwt ($\varphi$ =0)

2) From right extreme (t=0, x=+A)

x= A sin (wt + π/2)

x= A coswt

3) From left extreme (t=0, x=-A)

 x= A sin(wt - $\pi$\2)

x= -Acoswt

4) from half of maximum displacement to A

x= Asin(wt +$\pi$/6) as,

$\varphi$ = sin^-1(x/a) where x = A/2

Time taken by the particle

1) from (x=0 to A/2)

x = A sinwt

or, A/2=Asinwt

$\therefore$t=T/12

 2)from (x=A to A/2)

x = A cosw t  

Or,A/2=Acoswt

 $\therefore$t=T/6

3)from one extreme to another extreme.(x=-Ato A)

t=T/2 

4)From mean position to extreme position,

x = A sinwt

T= T/4 

Where, T isthe total time period from A to-A and from -A to A.

Velocity in SHM

V= $\frac{dx}{dt}=Awcos(wt+\varphi )=w\sqrt{A^2-x^2}$

So, $V_{\text{max}}=Aw$at x=0

So, $V=Aw\sqrt{\frac{1-x^2}{A^2}}=V_{\text{max}}\ast \sqrt{\frac{1-x^2}{A^2}}$

and $V_{\text{min}}=0$at x = A

Also, v²=w² (A²-x²)

Or, $\frac{v^2}{w^2}+x^2=A^2$

Or, $\frac{v^2}{A^2{w}^2}+\frac{x^2}{A^2}=1$

which shows graph between velocity and displacement on SHM is elliptical.

Acceleration on SHM

•$a=\frac{dv}{dt}=-A{w}^{2}sin(wt+\varphi )=-w^{2}x$

So, graph between a and x is st. line.

•At x=0, a=0 (min), at v=A; a=-w²A.

• For maximum amplitude a₀ and maximum velocity V₀ Amplitude and angular velocity are given as $A=\frac{v_0^2}{a_0}$ $w=\frac{d_0}{v_0}$    so,$T=\frac{2\pi v_0}{a_0}$

Spring mass system

•for both horizontal and vertical arrangements  

F=-kx=ma

$a=\frac{-kx}{m}=-w^{2}x$

Or, $w=\sqrt{\frac{k}{m}}$

$T=2\pi \sqrt{\frac{m}{k}}$

•If mass of sparing is considered ($m_s$); $T=2\pi \sqrt{\frac{\frac{m+ms}{3}}{k}}$ 

Reduced Mass

It is the behaviour of combined two masses when they are attracted toward each other by a certain force.

**** Distance doesn't matters.

For two bodies joined as aside:

Time period is $T=2\pi \sqrt{\frac{\mu }{k}}$ where μ=reduced mass=$\frac{M_a.M_b}{M_a+M_b}$

Spring combination

Simple Pendulum

The time period of pendulum is  given by;

$T=2\pi \sqrt{\frac{R}{1+\frac{R}{L}}g}$

 for a very short pendulum; R=Radius of earth ;l<<<

so,$T=2\pi \sqrt{\frac{L}{g}}$

For $L\to \infty$; $T=2\pi \sqrt{\frac{R}{g}}$

For L=R, $T=2\pi \sqrt{\frac{R}{2g}}$ =59.8min.

Note time period is independent of mass of bob of pendulum.

Change on length

•If the length of pendulum increases, time period increases  as (Tα$\sqrt{l}$);

motion becomes slow and time with loose and vice versa.

(i)If initial time period is T1, and by some factor it changes to T₂ if (T2>T1)•Loss in time =$\frac{T_2-T_1}{T_1}$*86400 secs.

(ii)if (T2>T1)gain in time =$\frac{T_2-T_1}{T_1}$*86400 secs.

•If a simple pendulum is taken from mean sea level to height h, Loss in time = 13.5*h secs.(h in km)

•If a simple pendulum is taken from mean sea level to depth h, Loss in time =13.5*hsecs. (h in km)

•Due to change on temperature loss or gain=0.5αΔθ

Change in acceleration due to gravity

$T=2\pi \sqrt{\frac{l}{g_r}}$ where $g_r$ is the effective acceleration due to gravity.

In a lift

i. Moving upward: $g_r=g+a_0$

ii. Moving downward: $g_r=g-a_0$

iii. Moving with constant velocity: $g_r=g$

On inclined plane

$g=gcos\theta$ where $\theta$is angle made by inclined plane with horizontal

In a train

i. Accelerating with acceleration $a_0$: $g_r=\sqrt{g^2+a_0^2}$

ii. Moving with constant speed in circular path: $g_r=\sqrt{g^2+{\left(\frac{v^2}{r}\right)}^2}$

Tension on string

$T_{\text{max}}=\frac{m{v}^2}{r}+mg$

we have, $V_{\text{max}}=Aw=A\sqrt{\frac{g}{l}}$

so $T_{\text{max}}=\frac{m{\left(A\sqrt{\frac{g}{l}}\right)}^2}{r}+mg$

= $mg+mg\left(\frac{A^2}{l^2}\right)$

= $mg\left(1+\frac{A^2}{l^2}\right)$ where A is the amplitude

Energy in simple harmonic motion

P.E.=$\frac{1}{2}kx$=$\frac{1}{2}$$(mx{w}^2)\ast x$=$\frac{1}{2}m{x}^2{w}^2$=$\frac{1}{2}m{A}^2{w}^{2}si{n}^2\theta$

K.E.=$\frac{1}{2}m{v}^2$=$\frac{1}{2}{A}^2{w}^{2}co{s}^2\theta$=$\frac{1}{2}m{w}^2(A^2-x^2)$

T.E.=$\frac{1}{2}m{A}^2{w}^2$= constant

So, P.E α $x^2$α $si{n}^2\theta$ α $\frac{1-cos2\theta }{2}$ =$\frac{(1-cos2wt)}{2}$=$\frac{(1-cos2\pi ft)}{2}$

K.E α (A²-x²) α $co{s}^2\theta$= $\frac{(1+cos2\theta }{2}$=$\frac{(1+cos2wt)}{2}$=$\frac{(1+cos2\pi ft)}{2}$

Hence, K.E and P.E has double the frequency of SHM.

Examples

1. A body executes SHM under action of force F1 with frequency f₁. If force changed to so it executes SHM with frequency  f2 when both force simultaneously acts. what is frequency of oscillation?    

solution:

F=mrw² =$mr(2\pi \ast f{)}^2$ α f²

so,$F_1\alpha f_1^2$and , $F_2\alpha f_2^2$ 

 for, $F_1+F_2$$\propto$$(f_1^2+f_2^2)$$\propto f^2$ (say)                                               

or, $f_1^2+f_2^2=f^2$,f is required frequency.

1. A load of mass m falls from height h onto the scale pan hung from a spring of spring constant k and mass of scale pan is zero and mass m doesn’t bounce relative to pan, then amplitude of vibration is

Solution:

Initial Energy= Final energy

$mgh=-\frac{1}{2}k{x}^2+mgx$

$k{x}^2-2mgkx+2mgh=0$

so, $x=\frac{2mg\pm \sqrt{4m^2{g}^2+4\ast 2mgh\ast k}}{2k}$ 

Taking +ve as $4m^2{g}^2+..........$> mg

$x=\frac{mg}{k}\left(1+\sqrt{1+\frac{2hk}{mg}}\right)$