## A community for students. Sign up today

Here's the question you clicked on:

## yrelhan4 2 years ago System consists of two identical slabs each of mass m linked by compressed weightless spring of stiffness K as shown in figure. The slabs are also connected by a thread which is burnt at a certain moment. a) Find at what value of Delta ‘l’, the initial compression of spring, the lower slab will bounce up after the thread is burned through.

• This Question is Closed
1. yrelhan4

i got the following solution in my book. can anybody plz explain the energy conservation equation..

2. MrDoe

This is just applying the conservation of energy. We can equate the energy lost and gained by the system and since we know that energy is conserved this must be equal to zero. Therefore we break up each piece. −mg(x+Delta l) This part is describing the energy due to gravity taking into account any displacement. 1/2K(Delta l)^2 This describes the energy "released" by the spring as it extends (taking into account that this is based on delta l, the distance the spring extends). -1/2 kx^2 You should recognize this as the potential energy of any spring where x is the distance compressed from equilibrium. Again, it really boils to energy lost = energy gained

3. yrelhan4

@MrDoe initially the spring is compressed by delta l.. it rebounds to (x+ delta l) so i get the potential energy.. it comes back to its initial position, so i get the 1/2K(Delta l)^2 part.. now here's where the confusion is.. 'x' according to me is the extension in the spring after it passes its initial position.. shouldnt x be equal to delta l? i mean spring force is f= -kx.. directly proportional to displacement .. so it must be an SHM, right?

4. Diwakar

We are taking x and deltal as the displacements from natural length position. Yes, the block will execute SHM but deltal and x will not at all be equal becuase what are 'equals' in a SHM are the displacements from the equilibrium position and in a vertical spring oscillator , the eqm position IS NOT the natural length. Eqm position is the point where the net force is zero and the net force will be zero at x=mg/k. I am getting the value of deltal as 3mg/k. The distance from eqm position is therefore 2mg/k. And it will also jump above the height of 2mg/k from eqm position.i.e. mg/k from he natural lenght creating a restoring force of mg/k and hence pulling the other mass up.

5. yrelhan4

thank you!

6. Diwakar

you are welcome!!

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy