If the system is suspended by the mass m, the length of the spring is ℓ 1 . If it is inverted and hung by mass M, the length of the spring is ℓ 2 . Find the natural length of the spring.

Question

If the system is suspended by the mass m, the length of the spring is ℓ 1 ​ . If it is inverted and hung by mass M, the length of the spring is ℓ 2 ​ . Find the natural length of the spring.

TestHub · Accepted Answer

Let L be the natural length of the spring and k its force constant.

When the system is suspended by mass m, the spring carries the weight of M:

$k (l_{1} - L) = M g$

When inverted and hung by mass M, the spring carries the weight of m:

$k (l_{2} - L) = m g$

Dividing the two equations:

$\frac{l _{1} - L}{l _{2} - L} = \frac{M}{m}$

$m l_{1} - m L = M l_{2} - M L$

$L (M - m) \cdot (- 1) \Rightarrow L = \frac{m l _{1} - M l _{2}}{m - M}$

Hence option (C) is correct.

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