25. A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 8 ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant? Answer+ 1 S 4 1 S, x 2 =e-2 that is, the weight is approximately 0.14 ft below the equilibrium position. 13 4A MacBook Pro Λ & ☆ * + < ( 6 7 8 6 Q Search Web % <& 85 $4 E R T Y U O 0 1) D F G H J K L P , . V > C V B N M He command V

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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5.1 25 math differential equation
25. A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a
damping force that is numerically equal to the instantaneous velocity. The mass is initially released from
a point 1 foot above the equilibrium position with a downward velocity of 8 ft/s. Determine the time at
which the mass passes through the equilibrium position. Find the time at which the mass attains its
extreme displacement from the equilibrium position. What is the position of the mass at this instant?
Answer+
1
S
4
1
S, x
2
=e-2
that is, the weight is approximately 0.14 ft below the equilibrium position.
13
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Transcribed Image Text:25. A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 8 ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant? Answer+ 1 S 4 1 S, x 2 =e-2 that is, the weight is approximately 0.14 ft below the equilibrium position. 13 4A MacBook Pro Λ & ☆ * + < ( 6 7 8 6 Q Search Web % <& 85 $4 E R T Y U O 0 1) D F G H J K L P , . V > C V B N M He command V
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