A mass of 2 kg is hung from the lower end of a vertical spring and extends it by 40 cm. The mass is now pulled down a further 20 cm and is then released from rest so that it oscillates about the equilibrium position. Determine :
a) the spring stiffness constant k for the spring. (2 marks)
b) the time period of the oscillations. (4 marks)
c) the speed and acceleration of the mass when it is 15 cm from the equilibrium position. (6 marks)
d ) the maximum speed and maximum kinetic energy of the mass.
(4 marks)
e) the maximum accelerating force on the mass. (4 marks)
These are all standard formulas. What is it you do not understand?
To answer the given questions, we'll use the concepts of Hooke's Law and simple harmonic motion.
a) To find the spring stiffness constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, the mass of 2 kg extends the spring by 40 cm, so the displacement (x) is 0.4 m. The force (F) can be calculated using the equation F = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, F = 2 kg * 9.8 m/s^2 = 19.6 N.
Using Hooke's Law, we can rewrite the equation as -kx = 19.6 N. Thus, k = -F/x = -19.6 N / 0.4 m = -49 N/m.
The spring stiffness constant (k) for the spring is 49 N/m.
b) The time period of oscillations can be found using the formula T = 2π√(m/k), where T is the time period, m is the mass, and k is the spring constant.
In this case, T = 2π√(2 kg / 49 N/m). Let's calculate the value:
T = 2π√(2 / 49) ≈ 2.89 s.
The time period of the oscillations is approximately 2.89 seconds.
c) The speed (v) and acceleration (a) of the mass can be determined using the equations of motion for simple harmonic motion. At any point in SHM, the displacement (x) can be given by x = A * cos(ωt), where A is the amplitude and ω is the angular frequency (ω = 2π / T).
In this case, the amplitude (A) is 0.15 m (15 cm = 0.15 m), and the time period (T) is 2.89 s. Therefore, ω = 2π / T = 2π / 2.89 ≈ 2.18 rad/s.
To find the speed and acceleration, we can differentiate the displacement equation with respect to time.
v = dx/dt = -A * ω * sin(ωt).
a = d²x/dt² = -A * ω² * cos(ωt).
Substituting the values, we have:
v = -0.15 m * 2.18 rad/s * sin(2.18t).
a = -0.15 m * (2.18 rad/s)² * cos(2.18t).
When the mass is 15 cm from the equilibrium position, the displacement (x) is 0.15 m. We can calculate the speed (v) and acceleration (a) by plugging in this value into the derived equations, at t = 0.
v = -0.15 m * 2.18 rad/s * sin(2.18 * 0) = 0.
a = -0.15 m * (2.18 rad/s)² * cos(2.18 * 0) = -0.15 m * (2.18 rad/s)².
Therefore, when the mass is 15 cm from the equilibrium position, the speed is 0 m/s, and the acceleration is approximately -0.1365 m/s².
d) The maximum speed occurs when the displacement is maximum, which is equal to the amplitude (A). Therefore, the maximum speed (vmax) is equal to A * ω.
vmax = 0.15 m * 2.18 rad/s = 0.327 m/s.
The maximum kinetic energy (KEmax) can be found using the formula KEmax = 0.5 * m * vmax².
KEmax = 0.5 * 2 kg * (0.327 m/s)² ≈ 0.34 J.
The maximum speed of the mass is approximately 0.327 m/s, and the maximum kinetic energy is approximately 0.34 Joules.
e) The maximum accelerating force on the mass occurs when the displacement is maximum and equals Fmax = k * A.
Fmax = 49 N/m * 0.15 m = 7.35 N.
The maximum accelerating force on the mass is approximately 7.35 N.