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 answer these questions, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.
a) To find the spring constant, k, we can use the formula:
k = F / x,
where F is the force applied to the spring and x is the displacement from the equilibrium position.
We know that the force applied to the spring is equal to the weight of the mass, which is given by:
F = mg,
where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the mass is 0.30 kg, the force F exerted by the spring is:
F = (0.30 kg) × (9.8 m/s²) = 2.94 N.
The displacement from the equilibrium position is 0.150 m:
k = (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 conservation of mechanical energy.
The maximum potential energy of the mass is converted into maximum kinetic energy when the spring is released from its stretched position.
The potential energy stored in the spring is given by:
PE = (1/2)kx²,
where k is the spring constant and x is the displacement from the equilibrium position.
Given that the additional displacement of the spring is 0.100 m, the potential energy stored in the spring is:
PE = (1/2)(19.6 N/m)(0.100 m)² = 0.098 J.
Since the potential energy is converted into kinetic energy when the spring is released, the maximum kinetic energy of the mass is equal to 0.098 J.
The kinetic energy is given by:
KE = (1/2)mv²,
where m is the mass and v is the velocity.
Setting the potential energy equal to the kinetic energy:
0.098 J = (1/2)(0.30 kg)v².
Solving for v:
v² = (2 * 0.098 J) / (0.30 kg) = 0.6533 m²/s².
v = √(0.6533 m²/s²) = 0.809 m/s.
Therefore, the maximum velocity of the mass is 0.809 m/s.
c) The maximum acceleration of the mass can be determined using Hooke's Law and Newton's second law of motion.
According to Newton's second law, the force acting on an object is equal to the mass of the object multiplied by its acceleration:
F = ma.
In this case, the force acting on the mass is provided by the spring, and it is given by:
F = -kx,
where x is the displacement from the equilibrium position.
Setting both expressions for force equal:
-ma = -kx.
Rearranging the equation:
a = (k / m)x.
Given that k is 19.6 N/m, m is 0.30 kg, and x is the additional displacement of 0.100 m:
a = (19.6 N/m) / (0.30 kg) * (0.100 m) = 6.53 m/s².
Therefore, the maximum acceleration of the mass is 6.53 m/s².
d) The total mechanical energy of the mass is the sum of its potential and kinetic energy.
The potential energy was previously calculated as 0.098 J.
The kinetic energy can be calculated using the maximum velocity, v:
KE = (1/2)mv² = (1/2)(0.30 kg)(0.809 m/s)² = 0.098 J.
Therefore, the total mechanical energy of the mass is 0.098 J (which is equal to the potential energy).
e) The period, T, and the frequency, f, of the mass and spring system can be determined using the formula:
T = 2π√(m / k)
f = 1 / T.
Given that m is 0.30 kg and k is 19.6 N/m:
T = 2π√(0.30 kg / 19.6 N/m) = 2.91 s.
f = 1 / (2.91 s) ≈ 0.344 Hz.
Therefore, the period of the mass and spring system is 2.91 s and the frequency is approximately 0.344 Hz.
e) The equation for the time-dependent vertical position of the mass can be described by simple harmonic motion:
x(t) = A * cos(ωt + φ),
where x(t) is the displacement at time t, A is the amplitude (maximum displacement), ω is the angular frequency, t is time, and φ is the phase constant.
The angular frequency, ω, can be calculated using the formula:
ω = √(k / m).
Given that k is 19.6 N/m and m is 0.30 kg:
ω = √(19.6 N/m / 0.30 kg) ≈ 8.07 rad/s.
Assuming the initial displacement is 0 and the mass is released at t = 0, the phase constant φ is 0.
Therefore, the equation for the time-dependent vertical position of the mass is:
x(t) = A * cos(8.07 rad/s * t).
Note: The value of A (amplitude) is not provided in the given information, so it is not possible to determine the specific equation unless A is given or assumed.