Elastic String - help required

[ashokmittal]

A particle of mass m is suspended under gravity from a point of the ceiling by a light elastic string of natural length h. When it is in equilibrium the extension if the string is a. It is pulled down a further distance b, which is > a, and released. Show that when the string becomes slack the particle speed V is given by V^2 = g(b^2 – a^2)/a. Show that if b^2 > a(2h+a) it will hit the ceiling with a speed U given by U^2 = V^2 – 2gh.

Describe what happens i)if b<a, ii)if b>a, but b^2 < a(2h+a), iii) if b > a but the string is replaced by a spring


Any help would be great,

Thanks

[Andrew Mason]

A particle of mass m is suspended under gravity from a point of the ceiling by a light elastic string of natural length h. When it is in equilibrium the extension if the string is a. It is pulled down a further distance b, which is > a, and released. Show that when the string becomes slack the particle speed V is given by V^2 = g(b^2 – a^2)/a. Show that if b^2 > a(2h+a) it will hit the ceiling with a speed U given by U^2 = V^2 – 2gh.

Describe what happens i)if b<a, ii)if b>a, but b^2 < a(2h+a), iii) if b > a but the string is replaced by a spring
The first step is to find the spring constant. You know that a force of mg increases the length by a-h. So k(a-h) = mg; k= mg/(a-h)

Then use a potential energy approach. Initial potential energy of the string is:

\frac{1}{2}k(b-h)^2

String potential energy is converted into kinetic energy of the mass and gravitational potential energy of the mass.

The string begins to go slack when the string length is back to h.

So: KE_h + mg(b-h) = \frac{1}{2}k(b-h)^2

AM