IIT JEE , AIEEE Forum - Physics , Mechanics

tabish_iit

Q--SHOULD WE USE ENERGY CONSERVATION OR FORCE EQUATION IN CASE OF SPRINGS?
I HAD READ FROM INTERACTIVE PHYSICS(mtg) THAT ONE MUST USE ENERGY CONSERVATION FOR SPRINGS BECOZ ACCELERATION AT THE EXTREME POSITION IS NOT ZERO.
BUT IN PROBLEMS OF H.C.VERMA IN SIMPLE HARMONIC MOTION AND A FEW OTHER CHAPTERS IF I USE THE ENERGY CONSERVATION I GET WRONG ANSWERS,HOWEVER IF I USES FORCE EQUATION I GET THE RIGHT ANSWER?
PLEASE MAKE IT CLEAR WHERE I SHOULD USE ENERGY CONSERVATION AND WHERE FORCE EQUATION IN CASE OF A SPRING

mayur1412

hi..
as far as da problems on shm r concerned..! we shud actually u both..! energy conservation..n force eqns..1st apply energy conservation..n den convert da respect energy's into forces..label all da forces..actin on da body..n then convert into da respective SHM terms..n see if u get da answer/..!

n as far as da general spring problems r concerned..only energy conservation wud do the trik..!

if u find da ans convincin dont forget to rate me..!!
cheers..!

chirag

Force equation is a foolproof method....

But, by conservation of energy, you shld get your answer correctly... b'coz energy and force equation can be related to each by IInd or IIIrd kinematic eqn, where we can convert velocity in accln, vice-versa....

So,it doesn't matter, if you use either of them.....

But, for special cases, it may differ....

If you are convinced, please don't forget to rate me.....

Thanx

edison

Both the methods are equally good and applicable in solving spring related problems.

May be u are committing some fundamental mistake while solving such problems.

You may post the solution by different methods for solving same problem in which you are encountering the different results, so that we can suggest you the lacunae, if any.

Moreover chirag has rightly stated that both metnods converges to the same result and can be derived from the other.

tabish_iit

    I HAVE TRIED IT .
    U CAN UR SELF CONFIRM IT BY APPLYING THE 2 WAYS TO
    H C VERMA
    CHAPTER 8 Qno45
    CHAPTER 29 Qno 51
AND MANY MORE

abhijeet_0201

force eq is valid only at equilibrium position!!! remember this
therefore it is advised to use energy conservation!!!!!!!!!!!!!!!

abhijeet_0201

so energy conservation brings out the right ans in q45(since max disp is asked)
there is no chapter29 in part 1 .are u talking bout part 2???
butREMEMBER FORCE eq is always valid only at EQUILIBRIUM POSITION!!!!!!!!!!
thank u!!!!!!!!!!!!

cvramana

It all depends on what you want to find. If it is force or acceleration or velocity or momentum.

My method is use force when to find acceleration.
Use energy to find velocity.

tabish_iit

ok guys try this simple question

A VERTICAL SPRING (spring constant k) HAS A MASS (m) ATTACHED TO IT
AND THE SYSTEM IS IN EQLIBRM. FIND THE EXTENSION IN IT?
SEE THE 2 CASES

mg=kx

OR
mgx=(1/2)k(x^2)

and u get 2 diff ans??

plz explain

rohit_kshirsagar

I think using energy conservation is a better option unless u dont learn to select a suitable reference frame.
But once you are able to select a suitable frame both methods may be applied.

shine

hey tabish
we use mg=kx when we comprress or extend the spring very gently and the force works over it for a very short span
as otherwise we use the potential energy as 1/2 kx^2 and not just kx^2
the basic reason is that when foce works for a large time interval then some of the e.p.e stored i s used up in the form of heat
and we have only 1/2kx^2 left as e.p.e

if convinced plz rate me

sidlol

Tabish, your energy equation is completely wrong.

The concept behind this is that the total energy in an SHM remains constant.

We write the total energy as the sum of potential energy and kinetic energy.

For a horizontal spring: the total energy at any point

= 0.5mv² + 0.5kx²

Now, either you can follow the Interactive Physics rule of finding

directly, or you can differentiate this to be more formal.

Since E = constant.

dE/dt=0.

or mv dv/dt + kx dx/dt = 0.

Since dx/dt=v,

d²x/dt² + (k/m)x=0.

Now

²=(k/m).

or T=2*pi*

(m/k) .

The method is similar for vertical springs. You have to take initial extension and the gravitational potential energy into account.

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