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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»

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).



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