A massless spring with spring constant 18.8 N/m hangs vertically. A body of mass 0.400 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released.
How far below the initial position does the body descend?
What is the frequency of oscillation of the resulting motion, assumed to be simple harmonic?
What is the amplitude of the resulting motion?
How far below the initial position does the body descend?
What is the frequency of oscillation of the resulting motion, assumed to be simple harmonic?
What is the amplitude of the resulting motion?
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(a)
F = (m)(a) = -(k)(x)
F = (0.4 kg)(9.8 N/kg) = -(18.8 N/m) x
x = (0.4)(9.8) / -(18.8)
x = - 0.209 meters (descent downward)
(b)
T = 2π * √(m / k)
where:
T = period
m = mass
k = spring constant
T = 2π * √(0.4 / 18.8)
T = 2π * √(0.4 / 18.8)
T = 0.916 s
frequency = 1 / T
f = 1 / 0.916
f = 1.09 Hz
(c)
Amplitude = (1/2) (x)
A = (1/2) (0.209 m)
A = 0.105 m
Hope this helps...good luck!
F = (m)(a) = -(k)(x)
F = (0.4 kg)(9.8 N/kg) = -(18.8 N/m) x
x = (0.4)(9.8) / -(18.8)
x = - 0.209 meters (descent downward)
(b)
T = 2π * √(m / k)
where:
T = period
m = mass
k = spring constant
T = 2π * √(0.4 / 18.8)
T = 2π * √(0.4 / 18.8)
T = 0.916 s
frequency = 1 / T
f = 1 / 0.916
f = 1.09 Hz
(c)
Amplitude = (1/2) (x)
A = (1/2) (0.209 m)
A = 0.105 m
Hope this helps...good luck!