Question

Let T1 be the time period of a single spring mass system on a horizontal surface. Let T2 be the time period of the same system when hung from the ceiling. What is the expression for T1 & T2?

(a) T1 = T2 = 2π √(k/m)

(b) T1 = 2π √(m/k), T2 = 2π √(mg/k)

(c) T1 = T2 = 2π √(m/k)

(d) T1 = 2π √(m/k), T2 = 2π √(k/mg)

This question was posed to me in an interview for job.

This key question is from Oscillations in division Oscillations of Physics – Class 11

Answer 1

The correct option is (c) T1 = T2 = 2π √(m/k)

Explanation: For the spring mass system on the horizontal surface, F = -kx.

So, T1 = 2π/w = 2π √(m/k).                                                                                                      When the spring mass system is hung from the ceiling, at mean position mg = kx0.                                                      Let spring be pulled further by x, extension = x+x0,                                                                                            F = k(x+x0)-mg

= kx + kx0– mg = kx.                                                                                                                                                                              The force is proportional to x.                                                                                                                                        Thus, w = √(k/m) & T2 = 2π √(m/k).                                                                                                                                 Therefore, T1 = T2 = 2π √(m/k).