A block with mass M=3.4 kg is connected to two springs with force constants k1=5

Question

A block with mass M=3.4 kg is connected to two springs with force constants k1=50 N/m and k2=133 ~N/m The mass moves on a frictionless horizontal surface where it is displaced from equilibrium and released. In both cases below, the motion can be described not only by all the individual forces acting in the problem but also by an equivalent, or effective, spring constant keff. To find it use F=ma and make this equation look like this: md2xdt2=?(...)xmd2xdt2=?(...)x. Whatever ends up as (...)(...) is keff.

(a) What is the effective spring constant keff in (a)? N/m ( ± 2 N/m)

(b) What is the period of the motion in (a)? s ( ± 0.02 s)

(c) In the case shown as "b" in the figure, what is the effective spring constant keff? Hint: Both springs stretch when the block moves. As they do so, the two spring forces at point P must balance because if they didn't, the tiny amount of mass at point P would have a huge acceleration (a=F/m). N/m ( ± 2 N/m)

(d) What is the period of the motion in (b)? s ( ± 0.02 s)

Explanation / Answer

a) md2xdt2 = k1x - k2x = - (k1+k2)x

effective spring constant = k1 +k2 =183 N/m

b) PERIOD, T = 2pi sqrt(m/k) = 6.28 sqrt(3.4/183) = 0.856 s

c) Since they are in series, 1/keff = 1/k1 + 1/k2

    keff = k1*k2/[k1+k2] = 133*50/[133+50] = 36.34 N/m

d) T = 2pi sqrt(m/k) = 6.28 sqrt(3.4/36.34) = 1.92 s

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