A 1.92 kg horizontal linear oscillator in part of some old clock has a 0.017 kg

Question

A 1.92 kg horizontal linear oscillator in part of some old clock has a 0.017 kg mass pointer sitting on top of it (see diagram). The effective force constant of the applied springs connected to the mass is 20.0 N/m and the amplitude of the oscillations is 10.0 cm. a) What is the period of the oscillations? b) At what position is the pointer most likely yo fall off? Why there? If the pointer falls off at this "most likely position", how much, if at all, will that change c) amplitude, d) the energy, and e) the period?

Explanation / Answer

The system of an oscillator is given as,

Mass, m= 0.017 kg

Force constant, k= 20 N/m

Amplitude, A= 10 cm

a). Time period , T= 2m/k = 2x3.140.017/20

Or, T= 0.183 second

b). Displacement pf pointer, x= mg/k = 0.017x9.8/20

Or, x= 0.833 cm

c). New Amplitude, A' = A - x

Or, A' = 10 - 0.833

Or, A' = 9.167 cm

d). New energy of oscillating system is,

E = 1/2 kA'2 = 1/2×20x (0.9167)^2

Or, E = 8.403 joule

e). Time period will not change and remain the same,

T = 2m/k = 0.183 second.

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