A block of mass m = 12 kg is attached to one end of a massless, horizontal sprin

Question

A block of mass m = 12 kg is attached to one end of a massless, horizontal spring of spring constant k = 100 N/m. The other end of the spring is xed. The mass executes simple harmonic motion with a maximum displacement of xmax = 0.6 m from equilibrium. A quantity of putty of mass mp = 4 kg is then dropped
onto the oscillating block and adheres to it.

a) What are the initial period of oscillation T0 , amplitude of oscillation A0 , maximum speed v0,max, and total energy E0 , before the putty is dropped onto the block?
b) If the putty “collides” with the block in part (a) at the moment when the block is at its maximum displacement from equilibrium, what are the new period of oscillation T1 , amplitude of oscillation A1 ,maximum speed v1,max, and total energy E1 , after the putty is dropped onto the block?

c) If the putty “collides” with the block in part (a) at the moment when the block is passing through its equilibrium position, what are the new period of oscillation T2 , amplitude of oscillation A2 , maximum speed v2,max, and total energy E2 , after the putty is dropped onto the block?

Explanation / Answer

a)

y = 0.6sin(wt) with w = (k/m) = (100/12) --->
y = 0.6sin((100/12)t)

= (100/12) = 2.88675

T0 = 2/ = 2*3.14/2.88675 = 2.18 s

A0 = 0.6 m

v = 0.6(100/12)cos((100/12)t)
vmax = 0.6(100/12) m/s = 1.73 m/s

E0 = (1/2) * k A2 = 0.5*100*0.6*0.6 = 18.0 J
---------------------------------------------------------------

b)
y = 0.6sin((100/16)t)

= (100/16) = 2.5

T0 = 2/ = 2*3.14/2.5 = 2.51 s

A0 = 0.6 m

vmax = 0.6(100/16) = 1.5 m/s

E0 = (1/2) * m V2 = 0.5*16*1.5*1.5 = 18.0 J

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c)

= (100/16) = 2.5

T0 = 2/ = 2*3.14/2.5 = 2.51 s

the new max velocity is found by conservation of the momentum:
vmaxf *16 = vmaxi*12
vmaxf = vmaxi*12/16 = 1.732*12/16 = 1.299 m/s

The total energy of the system is the kinetic energy at vmax:
Etot = 1/2 mv^2 = 1/2*16*1.299^2 = 13.499 J

new amplitude is found by conservation of energy:
Ekinmax = Espring:
13.499 J = 1/2 kA^2 = 50A^2
A = 0.5196 m

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