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# Spring 1 has natural length 0.5 m and k1=25 N/m and spring 2 has natural length of 1 m and k2 = 10N/m. They are joined together and free ends are streched so that the ends are fixed to the two walls 2m apart as shown. The stretched lengths of spring 1 and 2 at equilibrium are?(springs are massless)

2 years ago

Susmita
425 Points

The initial length of the two springs is (0.5+1)=1.5 meter.
The final length is 2m.
So the total stretched length is (2-1.5)=0.5 m
They are subject to same force.So
F=kx=constant.
Or,k=1/x
So k1/k2=x2/x1
or,x2/x1=25/10=5/2
So x1=0.5×(2/(2+5))=0.5×(2/7)=1/7
And x2=0.5×5/7=5/14
So stretched length of spring1 is
(0.5+1/7)=9/14
And that of spring2 is
(1+5/14)=19/14
Hope this helps
2 years ago
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### Course Features

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