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Simple pendulum and acceleration due to gravity

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1. The problem statement, all variables and given/known data
A simple pendulum and a mass-spring system have the same oscillation frequency f at the surface of the Earth. The pendulum and the mass-spring system are taken down a mine where the acceleration due to gravity is less than at the surface. What is the change in the frequency of the simple pendulum and the change in the frequency of the mass spring system

Simple Pendulum ----------- Mass-Spring

A f increases ------------- f decreases
B f decreases ------------- f decreases
C f increases ------------- f stays unchanged
D f decreases ------------ f stays unchanged

2. Relevant equations
Potential Energy = mgh
Kinetic Energy = 1/2mv^2

3. The attempt at a solution

I thought that because the acceleration due to gravity was now lower that the new maximum potential energy would now be lower as it is dependent on the value of g which in now lower, and a lower maximum potential energy means a lower maximum kinetic energy. Therefore if the pendulum has to travel the same path but at a lower maximum speed then it should take longer so the frequency should be less. For the mass spring I assumed that the frequency would be unchanged as the lower speed it will gain travelling downwards is counteracted by the fact it loses speed slower on the way up so I would have given D as my answer. However the mark scheme gives C as the answer. Can someone explain why the frequency increases when the value of g is lowered?

Thanks
 

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What is formula of T for each ? Therefore , what is formula of f for each ?

Which one is dependent on g , and how ?
 
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So for the simple pendulum T = 2π √(l/g) and then T =1/f so f = (1/2π)√(g/l) therefore f is proportional to g for a pendulum with the same length. Wouldn't the frequency then decrease when g is decreased?

For the mass spring system the formula is independent of g so I can see why it has no effect there
 
So for the simple pendulum T = 2π √(l/g) and then T =1/f so f = (1/2π)√(g/l) therefore f is proportional to g for a pendulum with the same length. Wouldn't the frequency then decrease when g is decreased?

For the mass spring system the formula is independent of g so I can see why it has no effect there
I believe the answer should be D .
Consider the derivation of simple pendulum's SHM - torque at some instant t → τ ≈ mgl(θ) .

Restoring torque will decrease if g decreases . As a result , the time to finish one oscillation will increase , and hence , time period will increase , and frequency will decrease .
 
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CWatters

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So for the simple pendulum T = 2π √(l/g) and then T =1/f so f = (1/2π)√(g/l) therefore f is proportional to g for a pendulum with the same length. Wouldn't the frequency then decrease when g is decreased?

For the mass spring system the formula is independent of g so I can see why it has no effect there
I agree D. I think the book answer C is wrong.

I recall reading that gravity actually increases as you go down a mine because the density of the earth isn't uniform. Once you are something like half way to the centre then it starts falling. However that can't be the reason for the error because the problem statement says..

the mass-spring system are taken down a mine where the acceleration due to gravity is less than at the surface.
and problem statements take priority over reality in exams :-)

My guess is they modified a question that originally mentioned the period rather than the frequency.
 
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I recall reading that gravity actually increases as you go down a mine because the density of the earth isn't uniform. Once you are something like half way to the centre then it starts falling. However that can't be the reason for the error because the problem statement says..
Is this a fact ? Or do you mean to say that mass that affects the gravitational field decreases ?
 

haruspex

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Why? I know a formula where it gives relation with f.
It doesn't actually say the spring is vertical, but even supposing it is...
Over the oscillation distance, the force of gravity is effectively constant. That corresponds to a constant spring extension, additional to that involved in the SHM. The two forces cancel, so have no effect on the period.
I recall reading that gravity actually increases as you go down a mine
By PF's own DH? http://www.thenakedscientists.com/forum/index.php?topic=45321.0
 
The formula saying mass-spring executing S.H.M

T=2*pi*√(e/g)
What is e ? Also , although this is probably wrong , what conditions is this under ?
 
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Where e is? And can you post a link reference for this?

Here, e= elongation of the spring due to suspended mass.

I am sorry, I could not find any suitable link to show as a reference.
 
Here, e= elongation of the spring due to suspended mass.

I am sorry, I could not find any suitable link to show as a reference.
An example then ?
 
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Okay, I found this equation from my textbook. I was trying to upload the photo, and translate the others. The photo is not uploading. So, let me give the derivation. I will upload it later.

Suppose a spring is suspended vertically with a mass m, so that it
undergoes an elongation e, then we can write,

Upward tension on the spring
T=ke
which is equal to the weight of the mass.
k=spring constant.
So, mg=ke.

Now if the mass is pulled x' (x'<e), further and left, then it will undergo SHM with an amplitude x'.

Let at time t, it has an acceleration a, distance from equilibrium is x and upward tension then is T'.
So, T'=k(e+x)
So, net force F=ma
Or, mg-T'=ma
Or, mg-k(e+x)=ma
Or, mg-mg-kx=ma
Or, ma=-kx
Or, a = -(k/m)x=(omega)^2*x

Where, omega=angualar velocity = 2*pi/T
T=time period.

So, we get,
T=2 pi sqrt(m/k)
But we have mg=ke, or, m/k=e/g
So, T=2*pi*√(e/g)
 
So, we get,
T=2 pi sqrt(m/k)
But we have mg=ke, or, m/k=e/g
So, T=2*pi*√(e/g)
The original equation that you've got is made of terms totally independent of each other - the second does not .

Look at you second equation - if you say , suppose , increase g , e will increase too . Thus the terms inside the root aren't independent of each other .

I hope this helps .
 

haruspex

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Okay, I found this equation from my textbook. I was trying to upload the photo, and translate the others. The photo is not uploading. So, let me give the derivation. I will upload it later.

Suppose a spring is suspended vertically with a mass m, so that it
undergoes an elongation e, then we can write,

Upward tension on the spring
T=ke
which is equal to the weight of the mass.
k=spring constant.
So, mg=ke.

Now if the mass is pulled x' (x'<e), further and left, then it will undergo SHM with an amplitude x'.

Let at time t, it has an acceleration a, distance from equilibrium is x and upward tension then is T'.
So, T'=k(e+x)
So, net force F=ma
Or, mg-T'=ma
Or, mg-k(e+x)=ma
Or, mg-mg-kx=ma
Or, ma=-kx
Or, a = -(k/m)x=(omega)^2*x

Where, omega=angualar velocity = 2*pi/T
T=time period.

So, we get,
T=2 pi sqrt(m/k)
But we have mg=ke, or, m/k=e/g
So, T=2*pi*√(e/g)
Ok, now it makes sense. (By elongation here they mean the amplitude of the oscillation.)
The equation is correct, but somewhat misleading in the present context. As you go down the mine, both g and e change. What's constant are m and k. Hence your equation does not help answer this problem.
 
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Ok, now it makes sense. (By elongation here they mean the amplitude of the oscillation.)
The equation is correct, but somewhat misleading in the present context. As you go down the mine, both g and e change. What's constant are m and k. Hence your equation does not help answer this problem.

Not amplitude, here amplitude is x'.

Well, I understood why g has no effect. So, D maybe an amswer.
 
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It is the elongation made when a mass is hung with the spring. when mass and 'g' is constant, 'e' is also constant.
 

haruspex

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It is the elongation made when a mass is hung with the spring. when mass and 'g' is constant, 'e' is also constant.
You're right, my mistake. In fact, that has to be what it represents to make my comments in post #17 right.
 

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