A 0.11kg mass is vibrating at the end of a horizontal spring along a frictionless surface. If the spring constant is 0.52 N/m, what is the displacement of the mass when the acceleration of the mass is 0.25m/s^2?
Acceleration - -w^2 * (displacement)
The angular frequency of vibration, in radians/s, is sqrt(k/m) = 2.17 rad/s
Solve for the displacement, X
X = -(acceleration) * (m/k)
To find the displacement of the mass, we can make use of Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position, as long as the spring remains within its elastic limit.
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,
x is the displacement from the equilibrium position.
In this case, we are given the mass of the object (0.11 kg), the spring constant (0.52 N/m), and the acceleration of the mass (0.25 m/s^2). We need to find the displacement, x.
To begin, let's rearrange the equation for acceleration:
F = ma
Rearranging for force:
F = ma
Plugging in the given values:
F = (0.11 kg)(0.25 m/s^2)
F = 0.0275 N
Now we can use Hooke's Law to find the displacement.
0.0275 N = - (0.52 N/m) x
Rearranging the equation for displacement:
x = - (0.0275 N) / (0.52 N/m)
Calculating the displacement:
x = - 0.0529 m
Therefore, the displacement of the mass is -0.0529 m. The negative sign indicates that the displacement is in the opposite direction of the force applied by the spring.