Do Adults and Children Swing at the Same Frequency?

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SAT2400
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URGENT//Period and frequency question

Homework Statement


1) An adult and a child are sitting on adjacent identical swings. Once they get moving, the adult, by comparison to the child, will necessarily swing with
a) a much greater period
b) a much greater frequency
c) the same period
d) the same amplitude


Homework Equations


T= 2pi(square root of m/k)


The Attempt at a Solution



THe answer is a B... Can anyone explain why?? I think it's an A b/c as m increases, the T increases??!
 
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I recall the following:
For small angles of excursion (i.e. for cases when the pendulum is gently swinging over a few degrees) the period of a pendulum approximates: t = 2*pi*(sqrt(l/g)). There is no term in this equation for mass, which in itself suggests that the size of the swinging mass is not important in determining the period. (l is length of pendulum, g is acceleration due to gravity)
As for the amplitude of the swinging, this has to do with the amount of force used to start the swinging. If the adult is twice the mass of the child, then for the same amplitude of swinging twice as much force (F=m.a) is required.
 


Thank you for the reply...

The answer is B. Do you agree with this??

Some of my classmates think it's a C...

Could you please explain again why the answer is a B??

Thank you very much T_T
 


well, I'm a bit worried about the relation you have given for the period of the pendulum. Are you quite sure it's a swinging. non-elastic pendulum?

I suggest the following: take a short length of string and try the period of different masses.

You'll find that mass of pendulum makes no observable difference to the period. But there is what looks like a mass term in the relation you have given, and I wonder why. This leads me to worry that I haven't seen the whole picture. I don't want to get this wrong...

I wonder where you got the relation T=2pi(m/k)^(1/2) from?