Spring Harmonic Oscilation in Gravitational Field

[hi-bob]

1. A mass M is attached to a fixed spring of constant K and oscillating with an amplitude A, the spring is hung from the ceiling in a gravitational field with constant free-fall acceleration g.
2. What is the oscillation frequency? Is it affected by gravity?
Mx''=-Kx-mg => ??
3. I figured it should be simply \omega = sqrt(k/m), but it looks like I'm wrong. This question is part of a longer calculation in a multiple choice exam. The only solution I have for comparison is the final result of the computation, but I don't understand the physical explanation. From what they say the frequency is not sqrt(k/m)

 

[jgens]

Why would you assume that the "angular velocity" omega is equal to the frequency?

 

[rock.freak667]

From what they say the frequency is not sqrt(k/m)[/b]

At equilibrium

mg=kx => m/k = x/g

 

[hi-bob]

Why would you assume that the "angular velocity" omega is equal to the frequency?

When I said frequency I was actually referring to the angular velocity \omega=2pi*f.
They are practically equivalent.

At equilibrium

mg=kx => m/k = x/g

OK, how do I get information about the frequency/angular velocity from here?

 

[hi-bob]

So I asked around irl, looks like the answer sheet is wrong.
The oscillation frequency remains unchanged by gravity.

 

[rock.freak667]

So I asked around irl, looks like the answer sheet is wrong.
The oscillation frequency remains unchanged by gravity.

If

$$T =2 \pi \sqrt{\frac{m}{k}} \ and \ \frac{m}{k}=\frac{x}{g}$$

then doesn't

$$T = 2\pi \sqrt{\frac{x}{g}}$$


since T depends on g, doesn't g affect frequency,f (=1/T)?

 

[Doc Al]

since T depends on g, doesn't g affect frequency,f (=1/T)?

No, since x (the equilibrium displacement) is proportional to g. (x = mg/k) If x were a constant independent of g, then your reasoning would be correct.

 

[jgens]

If

$$T =2 \pi \sqrt{\frac{m}{k}} \ and \ \frac{m}{k}=\frac{x}{g}$$

then doesn't

$$T = 2\pi \sqrt{\frac{x}{g}}$$


since T depends on g, doesn't g affect frequency,f (=1/T)?

Certainly not. Think about it this way: Once I hang a spring in a gravitational field the string will strech until it reaches equilibrium at some distance x₀ and kx₀ = mg. At this point, the net force acting on the block of mass M is zero (just as if we suspended the block in the absence of a gravitational field). Since the net force on the block is zero, the only force that we need to consider is the additional restoritive force acting on the block when we strech the string some distance A from its equilibrium position at x₀. Hence, once the spring is at equilibrium our equation for the net force acting on the block becomes F = kx where x is the distance from the blocks equilibrium position at x₀. Because of this, the period of oscillation does not depend on the presence of the graviational field.

 

[RoyalCat]

If

$$T =2 \pi \sqrt{\frac{m}{k}} \ and \ \frac{m}{k}=\frac{x}{g}$$

then doesn't

$$T = 2\pi \sqrt{\frac{x}{g}}$$


since T depends on g, doesn't g affect frequency,f (=1/T)?

Try and look at T(g).

T(g) = 2π*√(x/g) = 2π*√(m/k) which is independent of g.

What is the ratio, x/g? It is m/k. And that is a constant value determined only by the spring and the mass of the object attached to it.

 

[rock.freak667]

I understand better now.

So even if you do this on the moon where g is different. x will extend in relation to g such that m/k will be the same to produce the same period then?

 

[Doc Al]

So even if you do this on the moon where g is different. x will extend in relation to g such that m/k will be the same to produce the same period then?

Exactly. The period has nothing to with gravity. Even if the spring and mass were oriented horizontally, the period of oscillation would be the same.

 

[rock.freak667]

Exactly. The period has nothing to with gravity. Even if the spring and mass were oriented horizontally, the period of oscillation would be the same.

thanks..I always used to get this type of question wrong even though I thought my reasoning was correct and no one ever told me the correct thing.