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berkeman

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Welcome to the PF.Dear All!

I have attached the question. Please explain as to why we are ignoring mg when we are taking equation of motion for the spring moving in the vertical axis?

Thank you

It must be part of the original problem statement. You are correct that if that system is oriented vertically in a gravitational field, there will be a force term due to gravity as part of the equation. Perhaps the problem stipulates that the system is oriented horizontally on a frictionless surface?

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jbriggs444

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So gravity just complicates the example without adding anything useful. Pretend the setup is horizontal on an air table if you like.

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AlephZero

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In real life, if the springs were actually a tightly stretched wire, the weight of the object would be negligible compared with the other forces acting on it. For example if you are studying the vibration frequency of a guitar string, you can ignore whether the orientation of the guitar is horizontal or vertical.

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HallsofIvy

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[tex]m\frac{d^2x}{dt^2}= -k(x- l)- mg[/tex]

which is the same as

[tex]m\frac{d^2x}{dt^2}= -kx- (kl+ mg)[/tex]

Notice that, without any motion at all, the gravitational pull on the mass would extend the length of the spring until the two forces offset: mg= k(x- l) or x= (mg+ kl)/k. If we took

[tex]m\frac{d^2x}{dt^2}= -k(x- (mg+kl)/k)= -kx- (kl+ mg)[/tex]

exactly what we had above.

In other words, adding gravity to such a problem is exactly the same as changing the 'natural length' of the spring from what we would see if the spring were lying on a horizontal surface to what it is hanging.

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AlephZero

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To linearize the problem we asssume the tension is constant, which is the same as ignoring change in length from ##l## to ##\sqrt{l^2 + y^2}## of the springs.

If the weight gives an equilibrium position at a non-zero angle ##\bar\theta##, the period for small oscillaations about that position is different by a factor involving ##\cos\bar\theta##.

But it is quite realistic for the tension T to be several orders of magnitude greater than the weight mg, so ##\bar\theta## is so small it (and the weight) can be ignored.

In "real world" vibration situations it is quite common for objects to have accelerations much bigger than 1g. For example a guitar string vibrating at 1000 rad/s (about 160 Hz) with an amplitude of 2mm peak to peak in the centre has an acceleration of about 100g. For practical purposes, the effect of an extra "1g" from the weight of the object is often negligible.

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sophiecentaur

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Thank you so much everyone! It makes sense now =)

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Could you please explain what you mean by the above line?..."

If the weight gives an equilibrium position at a non-zero angle ##\bar\theta##, the period for small oscillations about that position is different by a factor involving ##\cos\bar\theta##. "

.

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2Tsinθ - mg = ma

So how are oscillations varying by cosθ?

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