Home » Ask & Discuss » Physics. » Mechanics « Back to Discussion



Mechanics

Vivek  Roy's Avatar
Blazing goIITian

Joined: 20 May 2008
Post: 332
5 Sep 2008 11:57:33 IST
0 People liked this
3
1164 View Post
Find the maximum elongation of the spring
None

Find the maximum elongation of the spring


Share this article on:

Comments (3)

VARUN  RAJ's Avatar

Blazing goIITian

Joined: 16 Mar 2008
Posts: 1825
5 Sep 2008 12:41:37 IST
0 people liked this

THE MAXIMUM ELINGATION OF THE SPRING IS WHEN THE BLOCKS HV THE SAME VELCOITY
m1V=(m1+m2)V2
V2=M1V/M1+M2
SINCE NO EXT FORCE ACTS ON THE SYSTEM THE ENERGY WILL BE CONSERVED
SO LOSS IN K.E.=GAIN IN SPRING ENRGY
SOLVE AND GET THE ANSWER
PLS RATE ME IF U FIND ME USEFUL
!!!!CHEERS!!!!!!
anchit saini's Avatar

Blazing goIITian

Joined: 1 Feb 2008
Posts: 1251
5 Sep 2008 12:56:50 IST
0 people liked this

\mbox{Max compression would be when both the blocks move with common velocity =} \\ \\<br/>v_{com} =\frac{m_1v}{m_1 + m_2} \\ \\<br/>Initial \ energy \ = \frac{m_1v^2}{2} \\ \\<br/>= final \ energy =\frac{kx^2}{2} + \frac{m_1v_{com}^2}{2} +\frac{m_2v_{com}^2}{2} \\ \\<br/>On \ solving -> \\ \\<br/>x=\sqrt{\frac{m_1m_2v^2}{k(m_1+m_2)}}

anchit saini's Avatar

Blazing goIITian

Joined: 1 Feb 2008
Posts: 1251
5 Sep 2008 12:57:25 IST
1 people liked this

\mbox{Just an extension }->\\ \\<br/>\mbox{Irodov 1.152 - Two bars connected by a weightless spring }\\ \\ \mbox{of stiffness k and length (in the non-deformable state) }l_0 \\ \\ \mbox{ rest on a horizontal plane . A constant force F starts acting } \\ \\ \mbox{on one of the bars as shown in fig. Find max and min distances between } \\ \\ \mbox{the bars during the subsequent motion of the system} \\ \\  \\ \\<br/>\mbox{Here we go :D } \\ \\<br/>\mbox{Here relative velocity approach seems quite comfortable } \\ \\<br/>\mbox{At any time the distance between the blocks is} \\ \\<br/>l_0 + x \mbox{ where x is the elongation in spring} \\ \\<br/>\mbox{Equations of motion} -> \\ \\<br/>kx = m_1 a_1 \\ \\<br/>F - kx = m_2 a_2 \\ \\<br/>a_r= \mbox{rel. accn. of 2 with respect to 1}\\ \\<br/>=\frac{F-kx}{m_2} -\frac{kx}{m_1} \\ \\<br/>a_r=v\frac{dv}{dx}\\ \\<br/>-> \\ \\<br/>\int{vdv}=\int{[\frac{F-kx}{m_2} -\frac{kx}{m_1}] dx}\\ \\<br/>\mbox{Since initially v relative is 0 and finally also its 0 (cos both the blocks}\\ \\ \mbox{ with same velocity during max separation), the limits of L.H.S are from 0 to 0}\\ \\<br/>\mbox{while limits of R.H.S are from }l_0 \ to \ l_{max} \\ \\<br/>\\ \\<br/>\mbox{On solving , we get}-> \\ \\<br/>l_{max}=l_0 + \frac{2m_1F}{k(m_1+m_2)}




Quick Reply


Reply

Some HTML allowed.
Keep your comments above the belt or risk having them deleted.
Signup for a avatar to have your pictures show up by your comment
If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team
Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor


@ INR 1,995/-

For Quick Info

Name

Mobile No.

Find Posts by Topics

Physics.

Topics

Mathematics.

Chemistry.

Biology

Parents

Board

Fun Zone

Sponsored Ads