1) A 200 g object moves in simple harmonic motion such that its amplitude decrea

Question

1) A 200 g object moves in simple harmonic motion such that its amplitude decreases by 1.5% with each comple oscillation

a) If its initial amplitude is 15cm what will the amplitude be after 30 oscillations

b) How much energy has been lost by the oscillator during this time?

2) A dampled harmonic oscillator is made from a 0.8kg mass on a k=14N/m spring with a damping constant of 400g/s It is initially released at a distance of 18cm from equilibrium

a) what is the initial energy of the oscillator?

b) What is the negy of the oscillator after one oscillation?

c) How long does it take for the amplitude of the oscillator to drop to one fourth of its original amplitude

d) How long does it take for the energy of the oscillaotr to drop to one-fourth of its original enegry?

3) A 2.0 kg mass sits on top of a 9.0kg mass as they oscillate horizontally on a frictionless surface. The spring constant of the spring causing the motion is 80N/m and the coefficient of static friction between the two blocks is 0.40 What is the maximum oscillation amplitude for which the upper block does not slip?

Explanation / Answer

1)the second oscillation as a period equal to 1-0.015=0.985 times the amplitude of the first oscillation; the next oscillation has an amplitude that is 0.985 of the second one, which is 0.9058^2 of the first, therefore, the nth oscillation has amplitude:

A(n)=15.0(0.985)^(n-1)

so the 30st oscillation has amplitude:

A(30)=15.0(0.985)^29

3)

First you must find the equations of motion of the combined blocks. You would do this by solving the differential equation m*d

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