Q51(HCV). A block of mass m and having charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant k as shown in the figure. A horizontal electric field E parallel to the spring is switched on. Find the amplitude of the resulting SHM of the block.
Answer - qE/k
Some questions that I have regarding this question:
1. What does "unstressed" spring mean?
2. Why does this block undergo SHM, assuming "unstressed" meaning spring at its natural length?
Can someone explain a bit? Thanks. :)
262unstressed here means that initially the spring is at its natural length and hence no spring force .
due to the field E, a force q.E acts on the body hence it moves outwards causing the spring to stretch and apply a restoring force on the block.hence eventually the block executes shm.
the amplitude can be found easily by finding the distance between points where the force acting is max. and min
hence q.E=k.x
or, x=qE/k=A
30Now, I feel I had a rather stupid doubt. Thanks a lot for the explanation! [1]