A spring id vertically, as shown below, and a 500g mass is hung from its lower end. The EQUILIBRIUM extension of the spring is observed to be 10.0cm.
(I) What is the spring constant of the spring?
(II) If the is pulled down by 5.00cm and released, what is the frequency of the subsequent oscillations.
(i) The spring constant is
k = M*g/deflection
= 0.500 kg*9.8 m/s^2/0.10 m = 49 N/m
(ii) The oscillation frequency is
[1/(2 pi)]*sqrt(k/M)
and is independent of the deflection amplitude.
To answer both questions, we need to use Hooke's law, which states that the force exerted by a spring is directly proportional to its extension. The equation for Hooke's law is given by:
F = -kx
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement or extension from the equilibrium position.
(I) To find the spring constant:
Given that the equilibrium extension of the spring is 10.0cm or 0.1m, and a 500g mass is hung from the lower end, we can calculate the force exerted on the spring using the equation:
F = mg
Where:
m is the mass (0.5kg), and
g is the acceleration due to gravity (approximately 9.8m/s²).
F = (0.5kg)(9.8m/s²) = 4.9N
Using Hooke's law, we can substitute the values of F and x into the equation:
4.9N = -k(0.1m)
Rearranging the equation to solve for the spring constant k:
k = -4.9N / 0.1m
k = -49 N/m (Negative sign denotes that the spring pulls upward when extended)
Therefore, the spring constant of the spring is 49 N/m.
(II) To find the frequency of oscillation:
The frequency of oscillation can be calculated using the equation:
f = (1 / 2π)√(k / m)
Where:
k is the spring constant (49 N/m), and
m is the mass (0.5kg).
Substituting the values into the equation:
f = (1 / 2π)√(49 N/m / 0.5kg)
Simplifying:
f = (1 / 2π)√(98 N/kg)
f ≈ 0.312 Hz
Therefore, the frequency of the subsequent oscillations is approximately 0.312 Hz.