Amplitude of a mass joined to a spring in the presence of an E-field

AI Thread Summary
The discussion revolves around calculating the amplitude of simple harmonic motion (SHM) for a charged mass connected to a spring in an electric field. Participants highlight the importance of using the correct equations, particularly the force balance equation qE = kx, which applies at the extreme positions of the motion. There is confusion regarding the conservation of energy approach and the completeness of the problem statement, with suggestions to clarify the wording and context. The consensus is that the amplitude can be determined by equating the electric force to the spring force, emphasizing that this equation is valid only when the mass is at maximum displacement. Proper understanding of the problem setup and conditions is crucial for accurate calculations.
Shivang kohlii
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Homework Statement


A block of mass m having charge q placed on smooth horizontal table and is connected to a wall thorough an unstretched spring of constant k . A horizontal electric field E parallel to spring is switched on. Find the ampliture of the shm by the block.

Homework Equations


kx= qe
U of spring= 1/2kx[2][/SUP]

The Attempt at a Solution


Using conservation of energy and work done by field ,F = qEx
1/2kx^2 = qEx
X (amplitude ) = 2qE/x
But ans isn't this..
In answer directly the forces have been equated
qE = kx
.. I don't understand why has this equation been used to calculate amplitude when conservation of energy is used to calculate the maximum displacement of a body attached to spring when moved from equilibrium position in ?
 
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Your problem statement is very incomplete. Could you copy the exact wording ? The way it is formulated now the right answer could as well be the amplitude of the electric field: E !

Your conservation of energy equation is incomplete and F = qEx looks misleading.
 
BvU said:
Your problem statement is very incomplete. Could you copy the exact wording ? The way it is formulated now the right answer could as well be the amplitude of the electric field: E !

Your conservation of energy equation is incomplete and F = qEx looks misleading.

I don't understand how to write the conservation of energy equation for this case
 
BvU said:
Your problem statement is very incomplete. Could you copy the exact wording ? The way it is formulated now the right answer could as well be the amplitude of the electric field: E !
:smile: It is standard to minimize effort: if something is asked and something else is commented, you ignore the request and comment on the comment.
I realize you have difficulty with this exercise and I am trying to help.
Start with a clear problem description.
If you think you need energies, you should list some energy equations as well.

Is it clear to you what the situation is ? Do you have enough information ?
What is actually happening when E is switched on and what is asked of you (in particular: the amplitude of what?) ? Did you make a sketch ?
 
Shivang kohlii said:
In answer directly the forces have been equated
qE = kx
.. I don't understand why has this equation been used to calculate amplitude when conservation of energy is used to calculate the maximum displacement of a body attached to spring when moved from equilibrium position in ?
The problem has been correctly solved by equating ##qE=kx##.
This eqn applies only when the body is at the extreme positions, as then it has no acceleration ## (a)## there.
In all other positions in between, the eqn of motion is ##qE\pm ma=kx##.
 
What does Shivang have to say ?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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