Sir, Kindly upload about the Mechanism Combination of SPRINGS. And Sir You are really a genius teacher and you know the place where a student would get troubled and Thus you explain so beautifully that everything reaches deep in the mind and create a Permanent Effect there. Thank You Sir Once Again...... .
Sir,Would you please tell me that in angular harmonic motion Like the one you have explained in the video ..restoring torque "mglsinx" or restoring Force "mgsinx" works on the bob to bring it back to mean position..
sir time period ke derivation ko wihout using torque prove kr sakte h.because in reference books there used torque to prove this formula for time period
Sir spring constant 'k' is with negative sign..as when we exert force on spring... restoring force comes into play which moves spring in its initial position which is opposite in direction
sir pls provide solution of this question ; two pendulum of length 100cm and 110.25cm is oscillating in phase, after how many oscillation will they be in same phase
sir if we go to higher altitudes then g dec. and time period increases then their should be gain of time rather than loose of time . i really didn't get it ???????
sir when pendulum moves from left to right the acceleration will be in direction of displacement because force i.e mgsintheta is in direction of displacement so how simple harmonic motion is possible?
The angular displacement starts out negative on the left side, and proceeds to being positive on the right side. At the neutral position in the middle, the angular displacement is defined as zero. Initially, the angular displacement is negative and the angular acceleration is positive. At the end of the first half cycle, the displacement is positive and the angular acceleration is negative. Angular acceleration is proportional to and the negative of, the angular displacement. As a point of semantics, there is centripetal acceleration of a pendulum that coexists with the tangential acceleration (i.e. length * angular acceleration in radians/sec^2). So there really is no position where the acceleration is zero, in a pendulum's motion. But when we focus exclusively on the tangential acceleration, or exclusively on the angular acceleration, we see simple harmonic motion. The centripetal acceleration just constrains it to follow the circular path.
sir what is the role of enery in oscillation i mean to say only restoring force and inertia of motion is responsible for oscillation and there is no role of energy
Displacement amplitude DOES matter, because we've only made an approximation to prove that a pendulum follows simple harmonic motion. The differential equation really is the following, where theta is in radians: d^2 theta/dt^2 + g/L*sin(theta) = 0 This cannot be solved in closed-form. You need to make infinite series expansions of sin(theta) to solve it. The simple solution we learn in introductory physics, makes the first order approximation of the infinite series to get sin(theta) = theta, which is only valid for small angles. To quantify "small", a cutoff to remember is that u to 22 degrees, you will keep the error in the period calculation below 1%. We ultimately are only approximating it as simple harmonic motion. To get a pendulum that really does follow simple harmonic motion regardless of amplitude, hang a string between two inverted cycloids, from the cusp between them. This is the solution to the famous tautochrone problem, where a particle moves along a path in an amount of time that is independent of amplitude.
sir at the time of coming towards mean position in pendulum the acceleration and displacement will be in same direction ......how it is shm......please explain.....
You are missing the fact that the displacement is defined as zero at the neutral position in the middle. When it swings from left to right on the left side, it starts with negative displacement and positive acceleration. When it swings on the right side, it has positive displacement and negative acceleration.
Dear sir, I have some doubt that you should clear it out. If theta is small then and the Bob is swinging from point a to point b then distance between point a to b is relatively small, so how force come into play here since velocity change between point a to be must be relatively smaller and accelerations must be neglected too thus there must be no question of force here???
A force has to be present if it is going to do anything other than move at a constant speed in a straight line. There are two forces acting on a pendulum: 1. the tension in the rod or string, that keeps it moving in a circular path 2. the gravitational force on the bob The component of the gravitational force on the bob, m*g*sin(theta), in the direction tangent to the path of the pendulum, is the restoring force that makes it swing back and forth if it starts elsewhere than the neutral position in the middle. The value of theta may be small, but not non-existent. It is small enough that we can approximate sin(theta) as theta, but not small enough that we can approximate sin(theta) as zero. Approximating sin(theta) as theta enables to solve the diffEQ in closed form as a simple harmonic oscillation. But approximating sin(theta) as zero, will defeat the purpose of the problem entirely.
+Almas Khadim time perios is inversely proportional to the square root of acceleration . In a downward accelerating elevator effective acceleration decreases .hence T ( time period ) will increase .
Not enough information. We need to know whether the pendulum is accelerating or not, and which direction. If the elevator accelerates upwards, the apparent value of g will increase, such that g=g0+a, where g0 is the 9.8 m/s^2 value when stationary, and a is the acceleration of the elevator cab. If the acceleration of the elevator cab is downward, g still equals g0+a, except a becomes negative.
Hello Sir, I have a doubt in the chapter Oscillations. Question : Is the function y= sin^2t (^ denotes power) describes Simple Harmonic Motion? In the book they have given that it doesn't. But sin^wt=1/2-1/2cos2wt, which can represent a Simple Harmonic Motion, if we take the equilibrium position to be 1/2. Please help Sir.
You need to be more careful with your notation. The term "sin^2(t)" denotes the sine of t raised to the 2nd power. The parenthesis indicate the input of sine, and the "^2" indicates the exponent. I recommend writing this as "(sin(t)^2", when you have to write it in plain text, since this is what Google Calculator and Wolfram Alpha will understand without ambiguity. The term sin^wt is meaningless to me, because it is unclear what is the exponent, and what is the input to the sine function. It is true that: (sin(w*t))^2 = 1/2 - 1/2*cos(2*w*t) This could be seen as SHM with twice the period associated with w as the angular frequency. Squaring the trig functions usually indicates one of the energies involved in simple harmonic motion, such as the kinetic energy or the potential energy. Since you will be squaring either the position or the velocity, both of which are sinusoidal functions of time.
No. The reason we call it moment of inertia, is that moments in general mean a quantity multiplied by a distance from a reference point, and added up. In this case, the distance is squared, so it is a second moment of mass that is the equivalent of inertia for rotational mechanics. We simplify this term to moment of inertia, since we rarely care about the "first moment of inertia", and have nothing to confuse it with. Sometimes we call it "mass moment of inertia" to be specific, and avoid confusing it with another variant of the term I'll discuss below. There is a difference between the misnomered "moment of inertia" that is used in structural analysis, and the rotational inertia that is used in rotational mechanics. The first has nothing to do with inertia, and should really be called "second moment of area". It has a lot of properties in common with the mass moment of inertia, but it doesn't depend on mass or inertia at all. Instead of adding up mass*r^2, we add up area*r^2. So it is a second moment of area. Some call it the "Area Moment of Inertia". The area moment of inertia is independent of material, as it is a purely geometrical property. It is a measure of how strong in bending, a given beam's cross section geometry is, and is multiplied with Young's modulus to get the total flexural rigidity.
Simple harmonic motion takes the following form: d^2 x/dt^2 + K*x = 0 where x is a variable that indicates position (or angle if desired), and capital K is a constant that measures the ratio of restoring force to inertial property. The solution will come in the form of: x = A*cos(w*t) + B*sin(w*t) where A and B are constants that depend on initial conditions, and w=sqrt(K). Traditionally, we use a lowercase omega, instead of a w, but I'll type w for simplicity. It can also be written in the form of: x = xmax*cos(w*t + phi) The capital K is not necessarily the same thing as the spring constant, lowercase k, although it is related. In the mass-spring system, K as I define it, takes the place of k/m. For a simple pendulum, x gets replaced with theta, and K=g/L. For a physical pendulum, K=m*g*dcm/I.
sir if we take a pendulum high above the surface of the earth then g will decrease and so m will decrease and according to the equatuin if m decreases then T should also decrease.butt you are saying that T increases
m will not decrease. m*g will decrease when g decreases, but m remains the same because mass is conserved (relativistic effects neglected). If you take a pendulum to the moon, g will be about 1/6th of its value here on Earth. This means a 1 meter pendulum with a 1 kg bob that swings with a 2 second period, will swing with a period of 4.9 seconds on the moon. Side note: it is not just a coincidence that a 1 meter pendulum swings with a 2 second period, because an early idea of defining the meter was based on this concept, so our value of g is approximately pi^2 in meters/second^2. The bob will still be 1 kg on the moon, even if it weighs 1.6 Newtons instead of 9.8 Newtons. The bob's mass doesn't affect the swing period of the pendulum, because it contributes equally to the inertial property as it does to the restoring force.
sir u told that for small thita sin thita = thita....and also told that timetaken by the bob to come to its euilibrium is same erespective of the x .but sir if we increase x the thita will also increase and now sin thita is not equal to thita.how the bob will take same time to reach the equilibrium.....for example if we take the bob and drop it at an angle of 90○to the normal ,will it take same time if it is released from 30○?
Sir I aspire to become a teacher like you and deliver lectures as effectively as you do. Sir please give your blessings so that I too can mould minds and touch lives as you do.
thanks sir...u r so generous to upload these videos for students free of cost
Very good videos sir makes my base for doing questions
You are the legend in the history of physics
sir youre videos are just awesome and extremely interesting to study now i gain a new interest in physics
Pradeep sir i still cant believe it really don't depends on Y 😍soo interesting
Thnk u sir 4 clearing all my confusions📖🖋🖋🖋🖋💡💡
I wish my school teacher would have been like you.
U r an amazing teacher and makes concepts crystal clear
Sir include this video in oscillations and waves playlist.
Sir, Kindly upload about the Mechanism Combination of SPRINGS.
And Sir You are really a genius teacher and you know the place where a student would get troubled and Thus you explain so beautifully that everything reaches deep in the mind and create a Permanent Effect there. Thank You Sir Once Again...... .
Thank you sir,
Helped me a lot.!
First time watch his video just wow.....Thanks alot sir.....You are great.....🙏🙏🙏🙏
Sir,Would you please tell me that in angular harmonic motion Like the one you have explained in the video ..restoring torque "mglsinx" or restoring Force "mgsinx" works on the bob to bring it back to mean position..
sir time period ke derivation ko wihout using torque prove kr sakte h.because in reference books there used torque to prove this formula for time period
God bless you sir. This helped me a lot.
thankyouuuu sir 😊😊😊😊
thnku sir......ur way of explaining this oscillating chapter is very nice.......it makes us to learn amd understamd easily
Sir u didn't teach oscillation in spring when arranged parallel and in series.....pls do reply
Sir spring constant 'k' is with negative sign..as when we exert force on spring... restoring force comes into play which moves spring in its initial position which is opposite in direction
The value of k is positive. The equation F=-k*x, gets a negative sign attached in front of k.
sir compound pendulum ki definition mei vertical plane ka mtlb us plane mai oscillate Krna hota hai??
sir pls provide solution of this question ; two pendulum of length 100cm and 110.25cm is oscillating in phase, after how many oscillation will they be in same phase
sir if we go to higher altitudes then g dec. and time period increases then their should be gain of time rather than loose of time . i really didn't get it ???????
Amazing explanation sir there is no word
sir when pendulum moves from left to right the acceleration will be in direction of displacement because force i.e mgsintheta is in direction of displacement so how simple harmonic motion is possible?
The angular displacement starts out negative on the left side, and proceeds to being positive on the right side. At the neutral position in the middle, the angular displacement is defined as zero.
Initially, the angular displacement is negative and the angular acceleration is positive. At the end of the first half cycle, the displacement is positive and the angular acceleration is negative. Angular acceleration is proportional to and the negative of, the angular displacement.
As a point of semantics, there is centripetal acceleration of a pendulum that coexists with the tangential acceleration (i.e. length * angular acceleration in radians/sec^2). So there really is no position where the acceleration is zero, in a pendulum's motion. But when we focus exclusively on the tangential acceleration, or exclusively on the angular acceleration, we see simple harmonic motion. The centripetal acceleration just constrains it to follow the circular path.
Thank u very much sir for this video
How Time period dependent on length ...because u have taken length constant earlier
sir what is the role of enery in oscillation i mean to say only restoring force and inertia of motion is responsible for oscillation and there is no role of energy
Prakash Chandra Tripathy energy gives you an idea about the amplitude of oscillation . greater the energy , greater will be the amplitude of SHM .
I am proud those teacher who is helped student like u sir g..... Nd sir aap Gjb k teacher hoo... Nd Thnx
sir why string of the simple pendulum need to be flexible and inextensible
this is bcz in Theo.phys.we consider only ideal condition we also neglect air friction
You made the concept crystal clear,thanksss a lot
sir can i use this video for aipmt preparation or this is only of class 11 level
Sir i want to buy ur video lectures for class 12,
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you are really great sir,so keep it on
god bless you
sir,since we have proven that a simple pendulum is a SHM.That means T is also equal to 2π √y/a . So does displacement really matter or not ?
Displacement amplitude DOES matter, because we've only made an approximation to prove that a pendulum follows simple harmonic motion.
The differential equation really is the following, where theta is in radians:
d^2 theta/dt^2 + g/L*sin(theta) = 0
This cannot be solved in closed-form. You need to make infinite series expansions of sin(theta) to solve it. The simple solution we learn in introductory physics, makes the first order approximation of the infinite series to get sin(theta) = theta, which is only valid for small angles. To quantify "small", a cutoff to remember is that u to 22 degrees, you will keep the error in the period calculation below 1%. We ultimately are only approximating it as simple harmonic motion.
To get a pendulum that really does follow simple harmonic motion regardless of amplitude, hang a string between two inverted cycloids, from the cusp between them. This is the solution to the famous tautochrone problem, where a particle moves along a path in an amount of time that is independent of amplitude.
sir, every oscilatory motion is a periodic motion...shm is an oscilatory motion..but how then shm is not periodic?? plz ans sir
I am from pak you are the best teacher
sir are oscillation and harmonic motion same
Sir...why theta is equal to sin theta and tan theta....
thank you very much sir......
sir at the time of coming towards mean position in pendulum the acceleration and displacement will be in same direction ......how it is shm......please explain.....
You are missing the fact that the displacement is defined as zero at the neutral position in the middle. When it swings from left to right on the left side, it starts with negative displacement and positive acceleration. When it swings on the right side, it has positive displacement and negative acceleration.
Dear sir,
I have some doubt that you should clear it out.
If theta is small then and the Bob is swinging from point a to point b then distance between point a to b is relatively small, so how force come into play here since velocity change between point a to be must be relatively smaller and accelerations must be neglected too thus there must be no question of force here???
A force has to be present if it is going to do anything other than move at a constant speed in a straight line.
There are two forces acting on a pendulum:
1. the tension in the rod or string, that keeps it moving in a circular path
2. the gravitational force on the bob
The component of the gravitational force on the bob, m*g*sin(theta), in the direction tangent to the path of the pendulum, is the restoring force that makes it swing back and forth if it starts elsewhere than the neutral position in the middle. The value of theta may be small, but not non-existent. It is small enough that we can approximate sin(theta) as theta, but not small enough that we can approximate sin(theta) as zero. Approximating sin(theta) as theta enables to solve the diffEQ in closed form as a simple harmonic oscillation. But approximating sin(theta) as zero, will defeat the purpose of the problem entirely.
Why the tension is not resolved instead of weight
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Sir aap kaun se collage me padhate ho
Sir 11 वी 12 वी प्रायोगिक वीडियो अपलोड कीजिए plz
awesome...may god bless u sir
sir the time period of simple pendulum in an accelerated descending elevator will increase or decrease?plz reply sir
+Almas Khadim time perios is inversely proportional to the square root of acceleration . In a downward accelerating elevator effective acceleration decreases .hence T ( time period ) will increase .
Not enough information. We need to know whether the pendulum is accelerating or not, and which direction.
If the elevator accelerates upwards, the apparent value of g will increase, such that g=g0+a, where g0 is the 9.8 m/s^2 value when stationary, and a is the acceleration of the elevator cab. If the acceleration of the elevator cab is downward, g still equals g0+a, except a becomes negative.
What is the uses and importance of simple pendulum sir g
@@PradeepKshetrapal tnxxxx alllot sir
Hello Sir,
I have a doubt in the chapter Oscillations.
Question : Is the function y= sin^2t (^ denotes power) describes Simple Harmonic Motion?
In the book they have given that it doesn't. But sin^wt=1/2-1/2cos2wt, which can represent a Simple Harmonic Motion, if we take the equilibrium position to be 1/2. Please help Sir.
You need to be more careful with your notation. The term "sin^2(t)" denotes the sine of t raised to the 2nd power. The parenthesis indicate the input of sine, and the "^2" indicates the exponent. I recommend writing this as "(sin(t)^2", when you have to write it in plain text, since this is what Google Calculator and Wolfram Alpha will understand without ambiguity. The term sin^wt is meaningless to me, because it is unclear what is the exponent, and what is the input to the sine function.
It is true that:
(sin(w*t))^2 = 1/2 - 1/2*cos(2*w*t)
This could be seen as SHM with twice the period associated with w as the angular frequency.
Squaring the trig functions usually indicates one of the energies involved in simple harmonic motion, such as the kinetic energy or the potential energy. Since you will be squaring either the position or the velocity, both of which are sinusoidal functions of time.
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Sir k death ho gayi hai dost.
sir,
Is T=mgcosθ or T-mgcosθ=mv^2/r
I want to join your class Sir.
sir,
Is there any difference between moment of inertia and rotational inertia?
No. The reason we call it moment of inertia, is that moments in general mean a quantity multiplied by a distance from a reference point, and added up. In this case, the distance is squared, so it is a second moment of mass that is the equivalent of inertia for rotational mechanics. We simplify this term to moment of inertia, since we rarely care about the "first moment of inertia", and have nothing to confuse it with. Sometimes we call it "mass moment of inertia" to be specific, and avoid confusing it with another variant of the term I'll discuss below.
There is a difference between the misnomered "moment of inertia" that is used in structural analysis, and the rotational inertia that is used in rotational mechanics. The first has nothing to do with inertia, and should really be called "second moment of area". It has a lot of properties in common with the mass moment of inertia, but it doesn't depend on mass or inertia at all. Instead of adding up mass*r^2, we add up area*r^2. So it is a second moment of area. Some call it the "Area Moment of Inertia". The area moment of inertia is independent of material, as it is a purely geometrical property. It is a measure of how strong in bending, a given beam's cross section geometry is, and is multiplied with Young's modulus to get the total flexural rigidity.
sir your,s teaching is very good and helpful
sir, what is a differential equation of SHM ?
Simple harmonic motion takes the following form:
d^2 x/dt^2 + K*x = 0
where x is a variable that indicates position (or angle if desired), and capital K is a constant that measures the ratio of restoring force to inertial property.
The solution will come in the form of:
x = A*cos(w*t) + B*sin(w*t)
where A and B are constants that depend on initial conditions, and w=sqrt(K). Traditionally, we use a lowercase omega, instead of a w, but I'll type w for simplicity.
It can also be written in the form of:
x = xmax*cos(w*t + phi)
The capital K is not necessarily the same thing as the spring constant, lowercase k, although it is related. In the mass-spring system, K as I define it, takes the place of k/m. For a simple pendulum, x gets replaced with theta, and K=g/L. For a physical pendulum, K=m*g*dcm/I.
sir if we take a pendulum high above the surface of the earth then g will decrease and so m will decrease and
according to the equatuin if m decreases then T should also decrease.butt you are saying that T increases
m will not decrease. m*g will decrease when g decreases, but m remains the same because mass is conserved (relativistic effects neglected).
If you take a pendulum to the moon, g will be about 1/6th of its value here on Earth. This means a 1 meter pendulum with a 1 kg bob that swings with a 2 second period, will swing with a period of 4.9 seconds on the moon. Side note: it is not just a coincidence that a 1 meter pendulum swings with a 2 second period, because an early idea of defining the meter was based on this concept, so our value of g is approximately pi^2 in meters/second^2.
The bob will still be 1 kg on the moon, even if it weighs 1.6 Newtons instead of 9.8 Newtons. The bob's mass doesn't affect the swing period of the pendulum, because it contributes equally to the inertial property as it does to the restoring force.
i am very intrested in physics by following this
sir u told that for small thita sin thita = thita....and also told that timetaken by the bob to come to its euilibrium is same erespective of the x .but sir if we increase x the thita will also increase and now sin thita is not equal to thita.how the bob will take same time to reach the equilibrium.....for example if we take the bob and drop it at an angle of 90○to the normal ,will it take same time if it is released from 30○?
the most awesome teacher i ever seen
Sir I aspire to become a teacher like you and deliver lectures as effectively as you do. Sir please give your blessings so that I too can mould minds and touch lives as you do.
Yea it really nice feeling from student he can reach the heart of student
amazin teachin sir ...!!!!!!!!! no 1 can explain sooo..... deeply and so interestingly THANK YOU SIR
Sir you are god of physics
😇😇
is torsional pendulum not included in video
ashish sharma It's not there in NCERT.
Sir what is the meaning of SHM?
Please leave study
why Bob is chosen spherical for simple pendulum
Anmol Mahia it’s a little aerodynamic (kind of) as it is less affected by air resistance as compared to cubical or cuboidal boxes
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Thanks you sir. For uploading lecture video
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Thanks
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5:20 idk why i found it funny
c
Very nicely taught sir.......but the students in your class are utterly disappointing.
monu yadav
sir your academy teachers were good chemistry teacher but our school physics teacher Mr Prince was better than you
sir, Is T=mgcosθ or T-mgcosθ=mv^2/r
thanks sir
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Thanks sir
You are the bestest
Thanks sir