 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
d^{2}x/dt^{2} + ω^{2}x=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
This blog is a platform of Physics/maths/science for Engineering and Medical entrance examination like IITJEE,AIEEE,CBSE board exams.
SHM in short
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
To read an article about force on a conductor click the link given below.
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) ..."
Hope you like it
physics expert
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) ..."
Hope you like it
physics expert
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
E_{n}=n(σ/ε_{0}) (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).E_{0} (2)
where σ is the surface charge density of the conductor , (σΔS) is the amount of charge on the area element ΔS and E_{0} is the field in the region where charge element (σΔS) is located.
Now there are two fields present E_{σ} and E_{0} and the resultant field both inside and outside the conductor near area element ΔS would be equal to the superposition of both the fields E_{σ} and E_{0} . Figure below shows the directions of both the fields inside and outside the conductor
E_{in}=E_{0}=E_{σ}
Since direction of E_{σ} and E_{0} are opposite to each other and outside the conductor near its surface
E_{out}=E_{0}+E_{σ}=2E_{0}
Thus , E_{0} =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 E_{σ} 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,
E_{σ}=σ/2ε_{0} (6)
Now,
E=2E_{0}=2E_{σ}=(σ/ε_{0})n=E_{n}
which is in accordance with equation 1. Hence from equation 5
f=σ^{2}/2ε_{0} = (ε_{0}E^{2}/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.
E_{n}=n(σ/ε_{0}) (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).E_{0} (2)
where σ is the surface charge density of the conductor , (σΔS) is the amount of charge on the area element ΔS and E_{0} is the field in the region where charge element (σΔS) is located.
Now there are two fields present E_{σ} and E_{0} and the resultant field both inside and outside the conductor near area element ΔS would be equal to the superposition of both the fields E_{σ} and E_{0} . Figure below shows the directions of both the fields inside and outside the conductor
Now field E_{0} 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 E_{σ }on either side pointing away from the surfaceas shown in the figure given above. Since E=0 inside the conductor
E<sub>in=E_{0}+E_{σ}=0E_{in}=E_{0}=E_{σ}
Since direction of E_{σ} and E_{0} are opposite to each other and outside the conductor near its surface
E_{out}=E_{0}+E_{σ}=2E_{0}
Thus , E_{0} =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 E_{σ} 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,
E_{σ}=σ/2ε_{0} (6)
Now,
E=2E_{0}=2E_{σ}=(σ/ε_{0})n=E_{n}
which is in accordance with equation 1. Hence from equation 5
f=σ^{2}/2ε_{0} = (ε_{0}E^{2}/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
or
dV=σdT
where σ is the constant of proportinality and is known as thomson coefficent
π=Ts=T(dE/dT)
and σ =T(ds/dT)=T(d^{2}E/dT^{2})
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 v_{1} to v_{2} during the displacement d then from equation of motion with constant acceleration we have
v_{2}^{2} v_{1}^{2}=2ad or
a=v_{2}^{2} v_{1}^{2}/2d Using this acceleration in Newton's second law of motion
we have
F=m(v_{2}^{2} v_{1}^{2})/2d
or
Fd=m(v_{2}^{2} v_{1}^{2})/2
or
Fd=mv_{2}^{2}/2 mv_{1}^{2}/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 mv^{2}/2 is called the kinetic energy of the body
Thus
K=mv^{2}/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 (mv^{2}/2) arises purely becuase of the motion of the body
 In equation 7 quantity
K_{2}=mv_{2}^{2}/2
is the final KE of the body and
K_{1}=mv_{1}^{2}/2
is the initial KE of the body .Thus equation 7 becomes
W=K_{2}K_{1}=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
K_{f}=K_{i}+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 workkinetic 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
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