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### SHM in short

• Simple harmonic motion is simplest form of oscillatory motion
• SHM is a kind of motion in which the restoring force is propotional to the displacement from the mean position and opposes its increase.Mathematically restoring force is
F=-Kx
K=Force constant
x=displacement of the system from its mean or equilibrium position
Diffrential Equation of SHM is
d2x/dt2 + ω2x=0

• Solutions of this equation can both be sine or cosine functions .We conveniently choose
x=Acos(ωt+φ)
where A,ω and φ all are constants
• Quantity A is known as amplitude of SHM which is the magnitude of maximum value of displacement on either sides from the equilibrium position
• Time period (T) of SHM the time during which oscillation repeats itself i.e, repeats its one cycle of motion and it is given by
T=2π/ω
where ω is the angular frequency
• Frequency of the SHM is the number of the complete oscillation per unit time i.e, frequency is reciprocal of the time period
f=1/T
Thus angular frequncy
ω=2πf
• Total energy remains constant in a SHM.So you can find the energy at any position and differentiate it to find the out the frequency
• Problem of SHM are basically to find out the timeperiod.So the concenteration should be on getting the net restoring force
• The basic approach to solve such problem is
1. Consider the system is displaced from equilibrium position
2. Now consider all the forces acting on the system in displaced position
3. find the restore force which comes out to be in the form
4.F=-kx

### Vector Algebra 2(quick recap)

Blog for graduate level physics: Vector Algebra 2: "In this post we'll lern Vector algebra in component form. Component of any vector is the projection of that vector along the three coordinat..."

### Vector Algebra 1(quick recap)

Blog for graduate level physics: Vector Algebra 1: " Here in this post we will go through a quick recap of vector algebra keeping in mind that reader already had detail knowledge and problem s..."

### Blog for graduate level physics: Force on a conductor

Blog for graduate level physics: Force on a conductor: "We have already learned in our previous discussion that field inside a conductor is zero and the field immidiately outside is En=n(σ/ε0) ..."
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### Force on a conductor

We have already learned in our previous discussion that field inside a conductor is zero and the field immidiately outside is
En=n(σ/ε0)                               (1)
where n is the unit normal vector to the surface of the conductor. We also know that any charge a conductor may carry is distributed on the surface of the conductor.
In presence of an electric field this surface charge will experience a force. If we consider a small area element ΔS of the surface of the conductor then force acting on area element is given by
ΔF=(σΔS).E0       (2)
where σ is the surface charge density of the conductor , (σΔS) is the amount of charge on the area element ΔS and E0 is the field in the region where charge element (σΔS) is located.
Now there are two fields present  Eσ and E0 and the resultant field both inside and outside the conductor near area element  ΔS would be equal to the superposition of both the fields  Eσ and E0 . Figure below shows the directions of both the fields inside and outside the conductor

Now field E0 has same value both inside and outside the conductor and surface element ΔS suffers discontinuty because of the charge on the surface and this makes field  Eσ on either side pointing away from the surfaceas shown in the figure given above. Since E=0 inside the conductor
E<sub>in=E0+Eσ=0
Ein=E0=Eσ
Since direction of  Eσ and E0 are opposite to each other and outside the conductor near its surface
Eout=E0+Eσ=2E0
Thus , E0 =E/2                      (3)
Equation (2) thus becomes,regardless of the of ΔF=½(σΔS).E                  (4)
From equation 4 , force acting per unit area of the surface of the conductor is
f=½σ.E                                       (5)
Here  is the  Eσ electric field intensity created by charge on area element ΔS at the point very close to this area element. In this region this area element behaves as infinite uniformly charged sheet hence we have,
Eσ=σ/2ε0                                                  (6)
Now,
E=2E0=2Eσ=(σ/ε0)n=En
which is in accordance with equation 1. Hence from equation 5
f2/2ε0 = (ε0E2/2)n                (7)
This quantity f is known as surface density of force. From equation 7 we can conclude that regardless of the sign of σ and hence direction of E , f is always directed in outward direction of the conductor.

### Thomson Effect

• Thomson effect is related to the emf that develops between two parts of the single metal when they are at different temperature

• Thus thomson effect is the absorption or evolution of heat along a conductor when current passes through it when one end of the conductor is hot and another is cold

• If two parts of the metal are at small temperature difference dT ,themn the electric potential difference is proportional to dT dV α dT
or
dV=σdT
where σ is the constant of proportinality and is known as thomson coefficent

• Peltier coefficent and thomson coeficent are related to thermopower according to following relations
π=Ts=T(dE/dT)
and σ =-T(ds/dT)=-T(d2E/dT2)

• We have seen that all the three effects are defined in terms of three coefficent namely seaback,peltier and thomson coefficent but the basic quantity is thermo-power which is the rate of change of thermo-emf with temperature
• ### Kinetic Energy

For full notes on Work , Energy and Power visit Physicscatalyst.com
• Kinetic energy is the energy possesed by the body by virtue of its motion
• Body moving with greater velocity would posses greater K.E in comparison of the body moving with slower velocity
• Consider a body of mass m moving under the influenece of constant force F.From newton's second law of motion
F=ma
Where a is the acceleration of the body
• If due to this acceleration a,velocity of the body increases from v1 to v2 during the displacement d then from equation of motion with constant acceleration we have
a=v22 -v12/2d Using this acceleration in Newton's second law of motion
we have
F=m(v22 -v12)/2d
or
Fd=m(v22 -v12)/2
or
Fd=mv22/2 -mv12/2           (7)
We know that Fd is the workdone by the force F in moving body through distance d
• In equation(7),quantity on the right hand side mv2/2 is called the kinetic energy of the body
Thus
K=mv2/2
• Finally we can define KE of the body as one half of the product of mass of the body and the square of its speed
• Thus we see that quantity (mv2/2) arises purely becuase of the motion of the body
• In equation 7 quantity
K2=mv22/2
is the final KE of the body and
K1=mv12/2
is the initial KE of the body .Thus equation 7 becomes
W=K2-K1=ΔK           (9)
• Where ΔK is the change in KE.Hence from equation (9) ,we see that workdone by a force on a body is equal to the change in kinetic energy of the body
• Kinetic energy like work is a scalar quantity
• Unit of KE is same as that of work i.e Joule
• If there are number of forces acting on a body then we can find the resultant force ,which is the vector sum of all the forces and then find the workdone on the body
• Again equation (9) is a generalized result relating change in KE of the object and the net workdone on it.This equation can be summerized as
Kf=Ki+W           (10)
which says that kinetic energy after net workdone is equal to the KE before net work plus network done.Above statement is also known as work-kinetic energy theorem of particles
• Work energy theorem holds for both positive and negative workdone.if the workdone is positive then final KE increases by amount of the work and if workdone is negative then final KE decreases by the amount of workdone