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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 2
Question
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 2 inches above the equilibrium position. Find the equation of motion. (Use g
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Physics
3 years
2021-08-05T18:29:51+00:00
2021-08-05T18:29:51+00:00 1 Answers
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Answers ( )
Answer:
the equation of motion is
Explanation:
Given that,
The weight attached to the spring is 24pounds
Acceleration due to gravity is 32ft/s²
Assume x is the string length, 4inches
convert the length inches to to feet = 4/12 = 1/3feet
From Hookes law , we calculate the spring constant k
k = W / x
k = 24 / (1/3)
k = 24 / 0.33
k = 72lb/ft
If the mass is displace from its equilibrium position by amount x
the differential equation is
Auxiliary equation is
Thus the solution is,
The mass is release from rest
x'(0) = 0
Therefore
x(t) = c₁ cos4 √6t
x(0) = -2inches
c₁ cos4 √6(0) = 2/12feet
c₁= 1/6feet
There fore, the equation of motion is