A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equilibrium point and released. Determine a). the spring constant, k, b). the maximum velocity of the mass, c. ) the maximum acceleration of the mass, d.) the total mechanical energy of the mass, e.) the period and frequency of the mass and spring, and e.) the equation of time dependent vertical position of the mass.
To determine the properties of the spring-mass system, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position.
a) To find the spring constant, k, we can use Hooke's Law equation:
F = -kx
Where F is the force exerted by the spring, x is the displacement from the equilibrium position, and k is the spring constant.
Given that the mass, m, is 0.30 kg and the displacement, x, is 0.150 m, we can find the force exerted by the spring:
F = mg
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values, we get:
F = (0.30 kg)(9.8 m/s^2) = 2.94 N
Since the force exerted by the spring is opposite to the displacement, we can write:
-2.94 N = -k(0.150 m)
Simplifying, we find:
k = 19.6 N/m
So the spring constant is 19.6 N/m.
b) To find the maximum velocity, we can use the equation:
v = √(2k/m) * A
Where v is the maximum velocity, k is the spring constant, m is the mass, and A is the amplitude (the additional displacement from equilibrium).
Given that the amplitude, A, is 0.100 m, we can calculate the maximum velocity:
v = √((2 * 19.6 N/m) / 0.30 kg) * 0.100 m
v = 2.59 m/s
Therefore, the maximum velocity of the mass is 2.59 m/s.
c) To find the maximum acceleration, we can use the equation:
a = (k/m) * A
Where a is the maximum acceleration, k is the spring constant, m is the mass, and A is the amplitude.
Plugging in the values, we get:
a = (19.6 N/m) / 0.30 kg * 0.100 m
a = 6.53 m/s^2
So, the maximum acceleration of the mass is 6.53 m/s^2.
d) The total mechanical energy (E) of the system is the sum of potential energy (PE) and kinetic energy (KE). At the maximum displacement, all the energy is in the form of potential energy, and at the equilibrium position, all the energy is in the form of kinetic energy.
PE = (1/2)kx^2
KE = (1/2)mv^2
At the maximum displacement, the potential energy is:
PE = (1/2)(19.6 N/m)(0.150 m)^2 = 0.441 J
At the equilibrium position, the kinetic energy is:
KE = (1/2)(0.30 kg)(2.59 m/s)^2 = 0.998 J
Therefore, the total mechanical energy is the sum of the potential and kinetic energy:
E = PE + KE = 0.441 J + 0.998 J = 1.439 J
So, the total mechanical energy of the mass is 1.439 J.
e) To find the period (T) and frequency (f) of the mass-spring system, we can use the equations:
T = 2π√(m/k)
f = 1/T
Plugging in the values, we get:
T = 2π√(0.30 kg / 19.6 N/m) = 0.907 s
f = 1/0.907 s ≈ 1.10 Hz
So, the period of the mass-spring system is approximately 0.907 seconds, and the frequency is approximately 1.10 Hz.
f) The equation of the time-dependent vertical position (y) of the mass can be described by the simple harmonic motion equation:
y(t) = A * cos(ωt + φ)
Where A is the amplitude, ω is the angular frequency (ω = 2πf), t is time, and φ is the phase constant.
In this case, the amplitude, A, is 0.100 m, the angular frequency, ω, is 2πf (where f is the frequency), and the phase constant, φ, is 0.
So, the equation of the time-dependent vertical position is:
y(t) = 0.100 * cos(2πf * t)
Where f is approximately 1.10 Hz.
This equation gives you the position of the mass at any given time during its oscillation.
To solve this problem, we'll use Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement of the spring.
a) The spring constant, k, can be determined using Hooke's Law:
F = k * x
Where F is the force, k is the spring constant, and x is the displacement.
Given that a mass of 0.30 kg stretches the spring by 0.150 m, we can calculate the force:
F = m * g
= 0.30 kg * 9.8 m/s^2
= 2.94 N
Using Hooke's Law, we can now determine the spring constant:
k = F / x
= 2.94 N / 0.150 m
= 19.6 N/m
Therefore, the spring constant is 19.6 N/m.
b) To find the maximum velocity of the mass, we can use the equation:
v_max = sqrt(k / m) * A
Where v_max is the maximum velocity, k is the spring constant, m is the mass, and A is the amplitude.
Given that the spring is stretched an additional 0.100 m, the amplitude A is:
A = x + x_extra
= 0.150 m + 0.100 m
= 0.250 m
Substituting the values:
v_max = sqrt(19.6 N/m / 0.30 kg) * 0.250 m
= sqrt(65.3) * 0.250
= 2.23 m/s
Therefore, the maximum velocity of the mass is 2.23 m/s.
c) The maximum acceleration of the mass is given by:
a_max = k / m * A
Substituting the values:
a_max = 19.6 N/m / 0.30 kg * 0.250 m
= 16.4 m/s^2
Therefore, the maximum acceleration of the mass is 16.4 m/s^2.
d) The total mechanical energy of the mass is given by:
E_total = 0.5 * k * A^2
Substituting the values:
E_total = 0.5 * 19.6 N/m * (0.250 m)^2
= 0.613 J
Therefore, the total mechanical energy of the mass is 0.613 J.
e) The period of oscillation, T, can be determined using the equation:
T = 2π * sqrt(m / k)
Substituting the values:
T = 2π * sqrt(0.30 kg / 19.6 N/m)
= 1.51 s
The frequency, f, is the reciprocal of the period:
f = 1 / T
= 1 / 1.51 s
≈ 0.663 Hz
Therefore, the period of oscillation is 1.51 s and the frequency is 0.663 Hz.
e) The equation of the time-dependent vertical position of the mass is given by:
y(t) = A * cos(2π * f * t)
Where y(t) is the vertical position at a given time t.
Substituting the values:
y(t) = 0.250 m * cos(2π * 0.663 Hz * t)
Therefore, the equation of the time-dependent vertical position of the mass is y(t) = 0.250 m * cos(2π * 0.663 Hz * t).