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Simple Harmonic Motion problem

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shalikadm
#1
Apr24-12, 12:51 AM
P: 63
1. The problem statement, all variables and given/known data


An elastic string is attached to A on a horizontal plane and a mass is attached to the other side of the rope.Then the rope is pulled to C and release.It engage in SHM from C to B and D to E and so on
natural length = l
extension = a
max end points = E & C
  1. periodic time of the SHM ?
  2. Center if oscillation ?
  3. amplitude of the SHM ?


2. Relevant equations


3. The attempt at a solution
I know that it doesn't engage SHM from B to D as the rope has not exceeded its natural length at that time...From B to D it travels at a constant velocity...As the periodic time of the oscillation is the time that it takes to complete an oscillation,I think it must be 2tEC,but not sure maybe its 2(tED+tBC).
Actually I don't have an idea about the center if oscillation whether its A or B and D.
And also amplitude.I think its DE or BC not AE as the SHM happens in only that area..
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gneill
#2
Apr24-12, 01:32 AM
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Presumably the question would like you to consider the end-to-end motion of the mass (between C and E) to comprise the SHM. If that is the case, then what strategy would you propose for finding the period?
shalikadm
#3
Apr24-12, 01:56 AM
P: 63
I think that the period is 2tEC
tEC=tAD+tDE
to find tAD we can use SUVAT equations.
and tDE=(∏/2)/ω

gneill
#4
Apr24-12, 02:10 AM
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P: 11,682
Simple Harmonic Motion problem

Quote Quote by shalikadm View Post
I think that the period is 2tEC
tEC=tAD+tDE
to find tAD we can use SUVAT equations.
and tDE=(∏/2)/ω
Okay, but what then is ω? If the problem is cast in the form of spring-driven SHM (with an elastic string), perhaps you need to incorporate springlike properties in your solution? Define the mass M and the spring constant k and you should be able to derive the period in terms of k, M, l, and a.
shalikadm
#5
Apr24-12, 02:31 AM
P: 63
here ω is the angular velocity (dθ/dt)
With F=ma and Hooke's law,I can get a function like
and calculate the period..any way the process is simple for me...But I want's to know,
periodic time of the SHM
Center if oscillation
amplitude of the SHM

Just please define them using letters in the image..

optional:
I'm doing this sum with a small use of differential calculus.Actually according to the Applied mathematics syllabus in our country...Any way..
in the syllabus we don't have k,we use λ(modulus of elasticity)
and hence k=λ/l
thanks
gneill
#6
Apr24-12, 10:43 AM
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P: 11,682
The period of anything with a cycle is the time to start from a given state and return to that state. In this case pick a point on the trajectory of the mass and (such as point C) and find an expression in terms of l, a, M, and k (or ##\lambda## if you prefer) for time for the mass to travel from C to E and back to C.

For the center of oscillation I would need to know if there is a particular definition used in your coursework. The one I'm familiar with pertains to physical pendulums and is related to the Radius of Gyration for a pivoted pendulum. Otherwise I would make the obvious choice and say that it corresponds to the average position of mass during a period (point A).

The amplitude of the oscillation should be obvious from the trajectory; what is the maximum displacement from the center of oscillation?
shalikadm
#7
Apr24-12, 11:09 AM
P: 63
Center of oscillation is defined as where acceleration is zero.So its B for C to B motion and D for E to D motion.
And the amplitude is DE for DE motion and BC for BC motion..
(as shown in examples of my coursework..)

I don't know what it will be to the whole journey..And I think that we can't take such thing as whole journey but taking the two SHMs(BC & DM) separately maybe the right way... Am I correct..
gneill
#8
Apr24-12, 11:31 AM
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P: 11,682
Quote Quote by shalikadm View Post
Center of oscillation is defined as where acceleration is zero.So its B for C to B motion and D for E to D motion.
And the amplitude is DE for DE motion and BC for BC motion..
(as shown in examples of my coursework..)

I don't know what it will be to the whole journey..And I think that we can't take such thing as whole journey but taking the two SHMs(BC & DM) separately maybe the right way... Am I correct..
I really can't say for sure. That would seem to define the center of oscillation as the entire stretch from B to D.
shalikadm
#9
Apr24-12, 11:45 AM
P: 63
Quote Quote by gneill View Post
I really can't say for sure. That would seem to define the center of oscillation as the entire stretch from B to D.
Can a center be a stretch ?
anyway..I have some another question..
We know at the center of oscillation the acceleration is zero (a=0)....
no acceleration means no resultant force..If so...Is the center of oscillation ,a equilibrium position (a balanced position) ?
gneill
#10
Apr24-12, 11:54 AM
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P: 11,682
Quote Quote by shalikadm View Post
Can a center be a stretch ?
anyway..I have some another question..
We know at the center of oscillation the acceleration is zero (a=0)....
no acceleration means no resultant force..If so...Is the center of oscillation ,a equilibrium position (a balanced position) ?
I suppose so. You can place the mass anywhere between D and B with no motion and it won't move. Only when you stretch past D or B will there be potential energy infused into the system.
shalikadm
#11
Apr24-12, 12:00 PM
P: 63
What about the last question I asked(Is the center of oscillation ,a equilibrium position (a balanced position) ?)..Its not about the question we discussed above..
gneill
#12
Apr24-12, 12:10 PM
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P: 11,682
Quote Quote by shalikadm View Post
What about the last question I asked(Is the center of oscillation ,a equilibrium position (a balanced position) ?)..Its not about the question we discussed above..
I answered above If the mass doesn't move if it's placed there, the net forces on it must be zero -- hence it's in equilibrium.
shalikadm
#13
Apr24-12, 12:34 PM
P: 63
So can u answer this...?

A mass is hanged by an elastic rope..So there's an extension in the rope and the mass is in its equilibrium(balanced) position(T=mg)...ok..
now the mass is pulled down to some length from that balanced position..Then releases...
Now it asks to find the time when it passes the equilibrium position...I want to know whether the T=mg position is a=0(the center of SHM) position or not ?
gneill
#14
Apr24-12, 01:14 PM
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P: 11,682
Quote Quote by shalikadm View Post
So can u answer this...?

A mass is hanged by an elastic rope..So there's an extension in the rope and the mass is in its equilibrium(balanced) position(T=mg)...ok..
now the mass is pulled down to some length from that balanced position..Then releases...
Now it asks to find the time when it passes the equilibrium position...I want to know whether the T=mg position is a=0(the center of SHM) position or not ?
I can answer it, but it is more important that YOU can answer it What's the net force acting on an object when it's "in equilibrium"? What's the acceleration of an object in equilibrium?
shalikadm
#15
Apr24-12, 01:19 PM
P: 63
got ya !
shalikadm
#16
Apr25-12, 01:58 PM
P: 63
(Sorry if I'm making trouble...)
Is center of oscillation of any SHM ,in equilibrium ?
We know that in the center of oscillation in SHMs of masses on a spring or a elastic string there's no any resultant force...so it in equilibrium..
But when we consider about the SHM of a simple pendulum or a mass that engage in uniform circular motion....there's a resultant force at their center of oscillation(Centripetal force)....
So pls make this clear...thanks !
gneill
#17
Apr25-12, 02:14 PM
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P: 11,682
Quote Quote by shalikadm View Post
But when we consider about the SHM of a simple pendulum or a mass that engage in uniform circular motion....there's a resultant force at their center of oscillation(Centripetal force)....
So pls make this clear...thanks !
In these cases there is no force component that will change the speed of the mass at the equilibrium point. Centripetal forces do not act along the line of motion.


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