

InterviewSolution
Saved Bookmarks
1. |
A light vertical spring is stretched by `0.2 cm` when a weight of `10g` is attached to its free end. The weight is further pulled down by `1cm` and released. Compute the frequency and maximum velocity of load. |
Answer» (i) Force constant of the spring `K = ("Restoring Force")/("Increase in length") = (mg)/("Increase in length")` `= (10^(-2)x9.8)/(2x10^(-3)) = 49 N m^(-1)` Frequency `f = (1)/(2pi) sqrt((K)/(m)) = (1)/(2pi) sqrt((49)/(10^(-2))) = (35)/(pi) Hz` (ii) amplitude of motion `(A) =` distance through which the weight is further pulled down `= 1cm` `V_(max) = A omega = 10^(-2) m x 70 rads^(-1) = 0.7 ms^(-1)` |
|