Find the magnitude of small oscillations

AI Thread Summary
The discussion centers on determining whether a bead on a rope, connected to a spring, undergoes simple harmonic motion (SHM) when displaced horizontally. Participants analyze the forces acting on the bead, including gravitational and spring forces, leading to the conclusion that the net force is proportional to x^3, suggesting the motion is not SHM. There is debate over the role of the spring constant in the equations derived, with some arguing it is essential for accurate representation of the forces. The final consensus is that the calculations need to include the spring's force to clarify the nature of the motion. The thread highlights the complexity of analyzing oscillatory motion in constrained systems.
prabhat rao
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Homework Statement




a rope is tied between 2 walls as shown.a bead of mass 'm' is on the rope as shown. it is constrained to move in the horizontal direction. it is tied to a spring of force constant 'k'- N/m. the spring is initially at its free length 'H'. the bead is displaced by a small displacement 'x' in the horizontal direction. does it execute SHM.If so find the magnitude of small oscillations?
no friction.


the figure is attached!






Homework Equations


T =2 pie/omega


The Attempt at a Solution


Consider the spring to make an angle q with the vertical
The mass in equilibrium in the y direction at all the times
Fsin q = mg
F (h/l)=mg
F = mgl/h
-Fcosq = f_restoring
-Fx/l = f_restoring
-mgl/hl *x =f_restoring
-mgx/h = f_restoring
-mgx/h = ma
ma+mgx.h = 0
a differential equation
omega = sqrt (g/h)
T = 2 pie * sqrt (h/g)
Now the answer is dimensionally correct

method 2

Since the force exerted by the spring is the vectorial sum of the forces along both the directions
F_y/(F_x) = tan q
-F_x= f_restoring = F_y/(tanq)
F_y intially is mg
f_restoring = -mgx/h
so this would be give

T= 2 pie *sqrt (h/g)

An amazing result independent of the spring constant of the force
A spring can only influence the motion along the direction of the spring

Is the solution?? if yes can anybody explain what it means

thank you
 

Attachments

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    Shm.jpg
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Hello,

I have a result which shows that the motion is NOT a simple harmonic moiton.
Because the net force on the bead is proportional to x^3.
This is just my opinion.


Regards
 
Can u please give your method
 
An amazing result independent of the spring constant of the force

This is the first sign something is wrong. Looking at your equations, I can't find anything that actually states the force the spring acts on the object (if it's there, it's certainly missing the spring constant
 
Is the motion simple harmonic??Please help me
 
I'll give you a hint

F_y intially is mg

Wrong! There's a spring pulling on it too
 
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}...
View full post »

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