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Grade: 12th pass
        
A light elastic string whose  natural length is a has one end fixed point O, and to the other end is attached a weight which in equilibrium would produce an extension e. Show that it the weight be let fall from rest at O, it will come to stay instantaneously at a point √(2ae + e2 ) below the position of equilibrium.
6 months ago

Answers : (2)

Ayush_Deep
120 Points
							
Hello friend , 
Assume the motion as SHM ,for which equilibrium position will be mean position.
And at extreme position it will come to stay for an instant.
Extreme position will be below the equilibrium position by a length equal to amplitude of SHM.
Now , by mechanical energy conservation find the velocity of block at equilibrium position( mean position) which will be equal to AW .
For an extension equal to e , we know that w will be √(g/e).
Velocity at equilibrium will be √2g(a+e) = A √(g/e)
Now by solving the equation we can find value of A which will be the required answer
Thanks and good regards.
2 months ago
Vikas TU
11498 Points
							
Dear student 
Hope your doubt gets cleared 
Please refer the solved examples of the below link 
2 months ago
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Course Features

  • 110 Video Lectures
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  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


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