- #1
randommanonea
- 7
- 0
A spring of mass M is suspended from the ceiling of a room. Find the extension in the spring due to its own weight if it has a spring constant of value k.
I am getting answer as \frac{Mg}{2k}, but the answer given in back of the book is \frac{Mg}{3k}. What I did was :
Let the natural (in un-stretched position) length of the spring be L. Let us consider an element of length dx at a distance x from the bottom of the spring. Then the spring constant of this small spring is k_x = \frac{L}{dx} k. Tension in this spring is T_x= \frac{x}{L} Mg (due to the weight of the spring below it, neglecting the weight of this small spring). Therefore, the extension in this small spring, dl = \frac{T_x}{k_x} = \frac{Mg}{k L^{2}} x dx. Therefore total extension, l = \frac{Mg}{k L^{2}} \int_{0}^{L} x dx = \frac{Mg}{2k}.
So, where I am wrong; or is the answer in the book wrong ?
I am getting answer as \frac{Mg}{2k}, but the answer given in back of the book is \frac{Mg}{3k}. What I did was :
Let the natural (in un-stretched position) length of the spring be L. Let us consider an element of length dx at a distance x from the bottom of the spring. Then the spring constant of this small spring is k_x = \frac{L}{dx} k. Tension in this spring is T_x= \frac{x}{L} Mg (due to the weight of the spring below it, neglecting the weight of this small spring). Therefore, the extension in this small spring, dl = \frac{T_x}{k_x} = \frac{Mg}{k L^{2}} x dx. Therefore total extension, l = \frac{Mg}{k L^{2}} \int_{0}^{L} x dx = \frac{Mg}{2k}.
So, where I am wrong; or is the answer in the book wrong ?