For the system shown in fig.1-38, the block has a mass of 1.5 kg and the spring

Question

For the system shown in fig.1-38, the block has a mass of 1.5 kg and the spring constant k = 8.0 N/m. Suppose that the block is pulled down a distance of 12 cm and released. If the friction force is given by -bv, where b = 0.23 kg/s, find the number of oscillations made by the block during the time interval required for the amplitude to fall to one-third of its initial value. A massless spring of force constant 19 N/m hangs vertically. A body of mass 0.20-kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released. Find (a) how far below the initial position the body descends, (b) Frequency, and c) the amplitude of the resulting motion, assigned to be simple harmonic. A 3.0-kg particle is in simple harmonic motion in one dimension and moves according the equation x =

Explanation / Answer

amplitude will drop like e^(-gamma t)

where gamma = b/2m

so if will drop to 1/3

e^(-gamma t) = 1/3

gamma t = ln(3)

t = ln(3)/gamma = ln(3)/(0.23/(2*1.5))= 14.33 s

T = 2 pi sqrt(m/k) = 2*pi*sqrt(1.5/8)=2.72

so # cycles = 14.33/2.72= 5.27 cycles

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