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SHM

Ian

Posted 14 May 2006 - 04:48 PM

Hello can someone please explain quesiton 4 in the 2003 paper for me please? I don't actually have the quesiton in front of me but i'll be able to ge tit on monday. If anyone can tell me beforehand that would be great.

Thanks

Ian

broughy

Posted 14 May 2006 - 06:09 PM

i'll have a shot at it once you post the question, can't really help you without it!

Jason Bourne

Posted 14 May 2006 - 06:21 PM

4. (a) Force exerted by spring = force due to gravity on mass.

Hence

 F=mg

(b) (i) Use Hookes Law  F=-ky to find the spring constant k(in N/m) from  mg=-ky

where y is the initial extension 40mm. The UNBALANCED force is then calculated from  F=ky
where y is the second extension 30mm.

(ii)  m \frac {d^2y} {dt^2} =-ky hence  \frac {d^2y} {dt^2} = -\frac {k} {m} y

hence  w=\sqrt {\frac {k} {m}} hence  T=\frac {2\pi} {w} = 2\pi \sqrt{\frac {m} {k}}

Ian

Posted 14 May 2006 - 08:51 PM

Thanks!

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