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A mass is suspended at the bottom of two springs in series having stiffness 10 N/mm and 5 N/mm. The equivalent spring stiffness of the two springs is nearly
  • a)
    0.3 N/mm
  • b)
    3.3 N/mm
  • c)
    5 N/mm
  • d)
    15 N/mm
Correct answer is option 'D'. Can you explain this answer?
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A mass is suspended at the bottom of two springs in series having stif...
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A mass is suspended at the bottom of two springs in series having stif...
Analysis:
When two springs are connected in series, the equivalent spring stiffness can be calculated by adding the reciprocals of the individual spring stiffnesses.

Solution:
Given:
Stiffness of the first spring, k1 = 10 N/mm
Stiffness of the second spring, k2 = 5 N/mm

The equivalent spring stiffness, keq can be calculated as follows:

1/keq = 1/k1 + 1/k2

Substituting the given values:

1/keq = 1/10 + 1/5

Simplifying the equation:

1/keq = (1+2)/10

1/keq = 3/10

Taking the reciprocal of both sides:

keq = 10/3 N/mm

Converting the units to N/m:

keq = (10/3) * (1000/1) N/m

keq = 10000/3 N/m

keq ≈ 3333.33 N/m (rounded to one decimal place)

Answer:
The equivalent spring stiffness of the two springs in series is approximately 3333.33 N/m, which is closest to option D.
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