question_answer

A block of mass m is connected to one end of a spring of spring constant k. The other end of the spring is fixed to a rigid support. Tile mass is released slowly so that the total energy of the system is then constituted by only the potential energy, then d is the maximum extension of the spring. Instead, if the mass is released suddenly from the same initial position, the maximum extension of the spring now is (g = acceleration due to gravity)

A)  \[\frac{mg}{k}\]                              

B)  \[2d\]

C)   \[\frac{mg}{3k}\]                                          

D)  \[4d\]

Correct Answer: B

Solution :

 Case 1: If block is released slowly, Given, spring constant = k When mass m is suspended from spring, then d extension is developed in spring. From law of conservation of energy \[mgh=\frac{1}{2}k{{d}^{2}}+mg(h-d)\] \[mgh=\frac{1}{2}k{{d}^{2}}+mgh-mgd\] \[mgd=\frac{1}{2}k{{d}^{2}}\] \[\Rightarrow \]               \[d=\frac{2\,mg}{k}\]                                     ?(i) \[\Rightarrow \]               \[k=\frac{2mg}{d}\] Case 2: Again if block is released suddenly, further, then again, from law of conservation of energy \[mgh=\frac{1}{2}k{{(d+y)}^{2}}+mg(h-d-y)\] \[mg(d+y)=\frac{1}{2}k{{(d+y)}^{2}}\] \[mg(d+y)=\frac{1}{2}\frac{2mg}{d}({{d}^{2}}+{{y}^{2}}+2dy)\] \[{{d}^{2}}+dy={{d}^{2}}+{{y}^{2}}+2dy\] \[\Rightarrow \] \[y(y+d)=0\] \[y=-d\] \[\therefore \] Total displacement = (2d)