When a 5kg mass is hung from a vertical spring,it streches by 25cm.(a)what mass must be supported by d spring in order to make d system vibrate wit a period of 1s.(b)if d amplitude of d motion is 5cm,determine d position nd d velocity of d mass 0.3s afta passing d equilibrium position on its way downward.(c)what is the restoring force on d mass@dis instant?
To answer part (a) of the question, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement.
In this problem, we need to find the mass that must be supported by the spring in order to make the system vibrate with a period of 1s. The period (T) is the time it takes for one complete cycle of vibration.
The formula for the period of vibration of a mass-spring system is:
T = 2π√(m/k)
Where T is the period, m is the mass, and k is the spring constant.
In this case, we know the mass of the object (5kg) and the displacement of the spring (25cm = 0.25m). We want to find the mass required for a period of 1s.
Substituting the known values into the formula and solving for k:
1 = 2π√(m/k)
k = (4π²m) / T² = (4 * π² * 5) / 1² = 20π²
Now, to find the mass that must be supported by the spring for a period of 1s, we can rearrange the formula for the period and solve for m:
m = (T² * k) / (4π²) = 1² * (20π²) / (4π²) = 5kg
So, a mass of 5kg must be supported by the spring to make the system vibrate with a period of 1s.
Moving on to part (b) of the question, we are given the amplitude (A) of the motion (5cm = 0.05m) and we need to determine the position and velocity of the mass 0.3s after passing the equilibrium position on its way downward.
For a mass-spring system undergoing simple harmonic motion, the general equations for position (x) and velocity (v) as a function of time (t) are:
x(t) = A * cos(ωt)
v(t) = -Aω * sin(ωt)
Where A is the amplitude, ω is the angular frequency, and t is the time.
The angular frequency (ω) is related to the period (T) by the equation:
ω = 2π / T
Since the period is 1s, the angular frequency is:
ω = 2π / 1 = 2π rad/s
Now we can substitute the known values to find the position and velocity 0.3s after passing the equilibrium position:
x(0.3) = 0.05 * cos(2π * 0.3) ≈ 0.05 * cos(0.6π) ≈ 0.05 * (-0.809) ≈ -0.04045m
v(0.3) = -0.05 * 2π * sin(2π * 0.3) ≈ -0.05 * 2π * sin(0.6π) ≈ -0.05 * 2π * (-1) ≈ 0.314 m/s
Therefore, the position of the mass 0.3s after passing the equilibrium position is approximately -0.04045m (downward direction), and the velocity is approximately 0.314 m/s (downward direction).
Finally, for part (c) of the question, we need to find the restoring force on the mass at this instant.
The restoring force (F_r) in a mass-spring system can be calculated using Hooke's Law:
F_r = -kx
Substituting the known values:
F_r = -(20π²) * (-0.04045) = 8π² * 0.04045 ≈ 1.0185 N
Therefore, the restoring force on the mass at this instant is approximately 1.0185 N.