A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that numerically e
qual to the instantaneous velocity. The mass is initially released from a point 1 ft above the equilibrium position with a downward velocity of 14 feet/second. Determine the time at which the mass passes through the equilibrium position
1 answer:
Answer:
the time at which it passes through the equilibrum position is:
t = 0.1 second
Explanation:
given
w= 4pounds
k(spring constant) = 2lb/ft
g(gravitational constant) = 10m/s² = 32ft/s²
β(initial point above equilibrum) = 1
velocity = 14ft/s
attached is an image showing the calculations, because some of the parameters aren't convenient to type.
You might be interested in
Given :
A cell of e.m.f 1.5 V and internal resistance 2.5 ohm is connected in series with an ammeter of resistance 0.5 ohm.
To Find :
The current in the circuit.
Solution :
We know, resistance of the ammeter is in series with the circuit.
So, total resistance is :
R = 2.5 + 0.5 ohm
R = 3 ohm
Also, e.m.f applied is 1.5 V .
Now, by ohm's law :
Therefore, the current in the circuit is 0.5 A.