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Hey guys

Im revising for my end of year exams and this was a question i got wrong in a test we had this year and unforunately there arent worked answers. I sort of get close but not quite there.

The answer is A, which i get, sorta

[tex] F = ma = 2kx [/tex]

As per Newtons third law (the hint) saying that the pulling and pushing force in turn create a pushing and pulling force, thus doubling the force. So using

[tex] s = ut +1/2 at^2[/tex]

Now the different masses will go different distances so if we take a ratio of masses, ie the smaller one going further we get

[tex]\frac{4x}{5} = \frac{kxt^2}{m}[/tex]

[tex] \frac{4}{5} = \frac{kt^2}{m}[/tex]

here I am trying to use t as period and thus get f via T = 1/f

Now this works fine, you rearange for t and you get the answer A. However one thing that puzzels me is: if this is supposed to be the period, surely we should consider one whole oscilation, that is to say the distance would infact be 8/5, which throws a spanner in the works. Any ideas?

Cheers

-G

Im revising for my end of year exams and this was a question i got wrong in a test we had this year and unforunately there arent worked answers. I sort of get close but not quite there.

The answer is A, which i get, sorta

[tex] F = ma = 2kx [/tex]

As per Newtons third law (the hint) saying that the pulling and pushing force in turn create a pushing and pulling force, thus doubling the force. So using

[tex] s = ut +1/2 at^2[/tex]

Now the different masses will go different distances so if we take a ratio of masses, ie the smaller one going further we get

[tex]\frac{4x}{5} = \frac{kxt^2}{m}[/tex]

[tex] \frac{4}{5} = \frac{kt^2}{m}[/tex]

here I am trying to use t as period and thus get f via T = 1/f

Now this works fine, you rearange for t and you get the answer A. However one thing that puzzels me is: if this is supposed to be the period, surely we should consider one whole oscilation, that is to say the distance would infact be 8/5, which throws a spanner in the works. Any ideas?

Cheers

-G

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