Simple Harmonic Motion Concept Page - 10

Example

Find resultant SHM equation of two superimposing SHMs

Example: Find the resultant amplitude due to superposition of two SHMs, x1=10sin(ωt+30∘$x_1=10sin(\omega t+{30}^{\circ }$)cm and x2$x_2$= 10cos$10cos$ (ωt+60∘$\omega t+{60}^{\circ }$)cm ?

Solution:
x1=10sin(wt+300)$x_1=10sin(wt+{30}^0)$
x2=10cos(wt+600)=−sin(wt−30)=sin(wt+150)$x_2=10cos(wt+{60}^0)=-sin(wt-30)=sin(wt+150)$
Thus, we get the two vectors of magnitude 10$10$ inclined at angles 30o${30}^o$ and 150o${150}^o$. The included angle between them is 120o${120}^o$.
Thus the resultant is 102+102+2(10)(10)cos1200−−−−−−−−−−−−−−−−−−−−−−−√=10cm$\sqrt{{10}^2+{10}^2+2(10)(10)cos{120}^0}=10cm$

Example

Resultant amplitude for two SHMs with given phase difference

Example: If A1=5A$A_1=5A$ and  A2=8A$A_2=8A$ and if (phase difference) ϕ=2π3$\varphi ={\displaystyle \frac{2\pi }{3}}$  what is the resultant amplitude of the particle?

Solution:
R=A21+A22+2A1A2cosϕ−−−−−−−−−−−−−−−−−−√$R=\sqrt{A_1^2+A_2^2+2A_1{A}_{2}cos\varphi }$
  A1=5A$A_1=5A$, A2=8A$A_2=8A$ , ϕ=2π3$\varphi ={\displaystyle \frac{2\pi }{3}}$
R=7A$R=7A$

Example

Resultant amplitude for two SHMs which are in phase or out of phase

Example: A particle is subjected to two simple harmonic motions of the same frequency and direction. The amplitude of the first motion is 4.0 cm and that of the second is 3.0 cm. Find the resultant  amplitude if the phase difference between the two motions is 0∘$0^{\circ }$ ?

Solution:
In such situation, amplitudes are added by vector method.
AR=(4)2+(3)2+2(4)(3)cosϕ−−−−−−−−−−−−−−−−−−−−−√$A_R=\sqrt{{\left(4\right)}^2+{\left(3\right)}^2+2\left(4\right)\left(3\right)\cos \varphi }$
   =25+24cos0∘−−−−−−−−−−−√$=\sqrt{25+24\cos 0^{\circ }}$
   =7$=7$ cm

Example

Find resultant of two SHMs which are in perpendicular directions

Example: Two linear simple harmonic motions of equal amplitude and frequency are impressed on a particle along x and y-axis respectively. The initial phase  difference between them is π2${\displaystyle \frac{\pi }{2}}$. What is the resultant path followed by the particle?

Solution:
As  one  is  along  x-  direction  and  other  along  y.
Because  of  initial  phase  difference π2${\displaystyle \frac{\pi }{2}}$  it  remains same throughout  the  motion   which   happens   when it  travels  along  a  circle.

Example

Relative motion between two SHM

Example:
Two simple pendulums of lengths 100$100$ cm and 196$196$ cm are in phase at the mean position at a certain time. If T$T$ is the time period of shorter pendulum, find the minimum time after which they will be again in phase.Solution:
T=2πlg−−√$T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$
T2=2πl1g−−√$T_2=2\pi \sqrt{{\displaystyle \frac{l_1}{g}}}$
T1=2π100g−−−−√$T_1=2\pi \sqrt{{\displaystyle \frac{100}{g}}}$
T2=2πl2g−−√$T_2=2\pi \sqrt{{\displaystyle \frac{l_2}{g}}}$
T2=2π196g−−−−√$T_2=2\pi \sqrt{{\displaystyle \frac{196}{g}}}$
T1T2=57${\displaystyle \frac{T_1}{T_2}}={\displaystyle \frac{5}{7}}$
∴(ω1−ω2)t=2π$\therefore ({\omega }_1-{\omega }_2)t=2\pi$
(2πT1−2πT2)t=2π$({\displaystyle \frac{2\pi }{T_1}}-{\displaystyle \frac{2\pi }{T_2}})t=2\pi$
t=7/2$t=7/2$
t=3.5sec$t=3.5sec$

Law

Condition on torque for simple harmonic motion

The work done by the torque shouldn't produce any dissipation of energy. As 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, Torque of restoring force in simple pendulum doesn't dissipate energy of the system.

Definition

State and use the condition on torque for simple harmonic motion

Another common example used to illustrate simple harmonic motion is the simple pendulum. This idealized system has a one end mass-less string suspended a mass m$m$ and the other end fixed to a stationary point. If the mass is displaced by a small distance, the angle moved is small.
The torque on the fixed point P is τ=Iα$\tau =I\alpha$
- mg sin  (t)L = mL^2 ''(t)$\text{- mg sin (t)L = mL^2 ''(t)}$
′′(t)+g/L(t)=0${}^{″}(t)+g/L(t)=0$
This has the same form as simple harmonic motion equation, x′′(t)−2x(t)$x^{″}(t){-}^{2}x(t)$, and so the solution is (t)=0cos(t−ϕ)$(t){=}_{0}cos(t-\varphi )$, the angular frequency is =(g/L)(1/2)$=(g/L{)}^{(1/2)}$.
It is interesting to note that the mass does not appear in this equation. This means that the frequency of the period only depends on the length of the string and the force of gravity. Pendulums with shorter strings will oscillate faster than pendulums with longer strings. And the same pendulum on the moon, where the force of gravity is 1/6th$1/6th$ that of the gravity on the Earth, will also take longer to oscillate.
We have glossed over one important aspect, in that this analysis is true only for small angles of theta. We had to make the approximation that sin is aproximately the same as which is true only for small angles.

Example

Understand and find equilibrium position in a given physical system

Equilibrium position in a SHM is a point where forces are in equilibrium. Displacement of the particle executing SHM is zero at equilibrium position.

Example: A ball of mass 2 kg$2kg$ from a spring oscillates with a time period
2 s$2s$. Ball is removed when it is in equilibrium position, then spring shortens by how much?

Solution:
The block executes SHM$SHM$ about mean position O. When the block is detached from the spring at point O, the spring again reaches  upto its
natural length.Let the spring shortens by x$x$  meter.
Thus    F=Mg=kx$F=Mg=kx$⟹x=Mgk$⟹x={\displaystyle \frac{Mg}{k}}$Also  time period of oscillation        T=2πMk−−−√$T=2\pi \sqrt{{\displaystyle \frac{M}{k}}}$2=2πMk−−−√⟹k=Mπ2$2=2\pi \sqrt{{\displaystyle \frac{M}{k}}}⟹k=M{\pi }^2$
∴x=MgMπ2⟹x=gπ2$\therefore x={\displaystyle \frac{Mg}{M{\pi }^2}}⟹x={\displaystyle \frac{g}{{\pi }^2}}$  meter

Example

Determine whether a certain physical system will perform SHM based on force analysis

For a simple pendulum force acting on it is:
mglsinθ=−Iα$mglsin\theta =-I\alpha$

For small angle sinθ=θ$sin\theta =\theta$ which makes it α=−kθ$\alpha =-k\theta$

Given system with a conservative force and small displacement will perform SHM.

Example

Condition on torque for SHM

Example:
In the figure shown, the springs are connected to the rod at one end and at the midpoint. The rod is hinged at its lower end. Find the condition for rotational SHM of the rod (Mass m$m$, length l$l$) to occur.Solution:
The rod performs SHM if

• the angular displacement of the rod is small (assumed)

• Restoring torque > Disturbing torque

Also for a small angular displacement we have 
y=l2ϕx=lϕ$y={\displaystyle \frac{l}{2}}\varphi x=l\varphi$
Applying the condition 
Restoring torque > Disturbing torque
kyl2cosϕ+kxlcosϕ>mgl2sinϕ$ky{\displaystyle \frac{l}{2}}cos\varphi +kxlcos\varphi >mg{\displaystyle \frac{l}{2}}sin\varphi$       
Taking ϕ$\varphi$  to be small...
kyl2+kxl>mgl2ϕ$ky{\displaystyle \frac{l}{2}}+kxl>mg{\displaystyle \frac{l}{2}}\varphi$
kl24ϕ+kl2ϕ>mgl2ϕ$k{\displaystyle \frac{l^2}{4}}\varphi +k{l}^2\varphi >mg{\displaystyle \frac{l}{2}}\varphi$
54kl2ϕ>mgl2ϕ${\displaystyle \frac{5}{4}}k{l}^2\varphi >mg{\displaystyle \frac{l}{2}}\varphi$
k>25lmg

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