A 1.2-kg block rests on a frictionless surface and is attached to a horizontal s

Question

A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constantk = 39N/m (see the figure) (Figure 1) . The block oscillates with amplitude 10 cm and phase constant ?=??/2. A block of mass 0.80 kg moves from the right at 1.7 m/s and strikes the first block when the latter is at the rightmost point in its oscillation. The two blocks stick together.

Part A

Determine the frequency of the resulting motion.

Express your answer using two significant figures.

.56

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Part B

Determine the amplitude of the resulting motion.

Express your answer using two significant figures.

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Part C

Determine the phase constant (relative to the originalt=0) of the resulting motion.

Express your answer using two significant figures.

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Figure 1 of 1

A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constantk = 39N/m (see the figure) (Figure 1) . The block oscillates with amplitude 10 cm and phase constant ?=??/2. A block of mass 0.80 kg moves from the right at 1.7 m/s and strikes the first block when the latter is at the rightmost point in its oscillation. The two blocks stick together.

Part A

Determine the frequency of the resulting motion.

Express your answer using two significant figures.

f=

.56

  Hz

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Incorrect; Try Again; 4 attempts remaining

Part B

Determine the amplitude of the resulting motion.

Express your answer using two significant figures.

A=   cm

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Part C

Determine the phase constant (relative to the originalt=0) of the resulting motion.

Express your answer using two significant figures.

?= rad

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Figure 1 of 1

A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constantk = 39N/m (see the figure) (Figure 1) . The block oscillates with amplitude 10 cm and phase constant ?=??/2. A block of mass 0.80 kg moves from the right at 1.7 m/s and strikes the first block when the latter is at the rightmost point in its oscillation. The two blocks stick together. Part A Determine the frequency of the resulting motion. Express your answer using two significant figures. f= .56 Hz SubmitMy AnswersGive Up Incorrect; Try Again; 4 attempts remaining Part B Determine the amplitude of the resulting motion. Express your answer using two significant figures. A= cm SubmitMy AnswersGive Up Part C Determine the phase constant (relative to the originalt=0) of the resulting motion. Express your answer using two significant figures. ?= rad SubmitMy AnswersGive Up Provide FeedbackContinue

Explanation / Answer

before the impact:
y = Acos(wt + phi)
w = ?(k/m)
w = ?(39/1.2) = 5.7 rad/s
y = 0.1cos(5.7t - pi/2)
find t at which the object is at the rightmost point:
0.1 = 0.1cos(5.7t - pi/2)
cos(5.7t - pi/2) = 1
5.7t - pi/2 = 0
t = 0.275 s

the velocity of both blocks directly after the impact is
v = (m1v1 + m2v2)/(m1 + m2)
v = 0.8*1.7/2 = 0.68 m/s

so the new oscillation has a velocity of 0.68 m/s at distance 0.1 from the equilibrium point at t = 0.275 s
the new w is
w = ?(39/2)
w = 4.42 rad/s

(a)
f = w/(2pi) = 4.42/(2pi) = 0.7 Hz

(b)
the new equation for the position and velocity is
y = Acos(wt + phi)
v = -Aw sin(wt + phi).............(1)
v = -4.42A sin(wt + phi).........(2)

From (1)

0.1 = A*cos(4.42*0.275 + ?)
0.1 = A*cos(1.2155 + ?)..................(3)

From (2)

0.68 = - 4.42A sin(4.42*0.275 + ?)
0.68 = - 4.42A sin(1.2155 + ?).........(4)

Divide (4) by (3)
6.8 = -4.42 tan(1.1255 + ?)
tan(1.1255 + ?) = -1.54
(1.1255 + ?) = 2.1456
? = 1.02 rad
Putting it in (3)
0.1 = A*cos(1.2155 + 1.02)
0.1 = 0.62 A
Amplitude, A = 0.162 m

(c)
?? = pi/2 - 1.02 = 0.55 rad

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