Equilibrium and Elasticity Equilibrium and Elasticity Physics Help Forum

 Dec 28th 2009, 05:37 AM #1 Junior Member   Join Date: Dec 2009 Posts: 11 String mass of the body=m Horizontally, the spring is deflected from its equilibrium position and then released, oscillated with a frequency f. Then, the same spring and the same body are suspended vertically, as in figure. Which one is correct? A) The frequency of movement of the spring, when the body is suspended vertically, doubled. B) The frequency of movement of the spring, when the body is suspended vertically, reduced to half. C) The frequency of movement of the spring, when the body is suspended vertically, remains the same. D) The frequency of movement of the spring, when the body is suspended vertically, is not related to the period of the movement, when the spring oscillates horizontally. I think is C) but I think that D) is also correct. Please help me Attached Thumbnails
 Dec 28th 2009, 05:45 AM #2 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 please tell me why are u thinking d is correct too? when u agree in c) same frequency, then will u not agree that time period=1/f wil also be same for both and so related by T1=T2?
 Dec 28th 2009, 08:10 AM #3 Junior Member   Join Date: Dec 2009 Posts: 11 Yes I will agree with that but the change of position doesn't affect anything? In the second case we have 2 forces actuating: grav force and elastic force In the first case we have 1 force: elastic force Does this change anything???
 Dec 28th 2009, 08:54 AM #4 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 why dont we try to calculate the second case?? in the second case, my first task is to find where is the eq position. mg=k x0, ie x0=mg/k is the equilibrium elongation. let spring be further displaced by y. so new equation of motion is m a= mg-k(y+x0) so, ma=mg-ky-kx0=-ky(since we had the relation kx0=mg at eq) note that this is just the shm eq. so w=root k/m hence frequency and T remains same.

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