Question:

A 2.5 kg mass is attached to a horizontal spring with a spring constant of 100 N/m. The spring is initially compressed by 0.2 meters. The system is released from rest, causing the mass to undergo simple harmonic motion.

a) Calculate the maximum acceleration experienced by the mass during the motion.

b) Determine the total mechanical energy of the system.

c) At what displacement does the mass have a speed of half its maximum speed?

d) Find the period and frequency of the motion.

Assume no energy is lost due to friction or air resistance.

Answer:

a) The maximum acceleration experienced by the mass can be calculated using Hooke's law and Newton's second law.

According to Hooke's law, the force exerted by a spring is given by:

F = -kx,

where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.

Since acceleration is given by a = F/m, we can rewrite the equation as:

a = (-kx)/m.

Substituting the provided values:

k = 100 N/m, x = 0.2 m, m = 2.5 kg,

we can calculate:

a = (-100 N/m * 0.2 m) / 2.5 kg = -8 m/s^2.

However, since the magnitude of acceleration is always positive, the maximum acceleration is 8 m/s^2.

b) The total mechanical energy of the system is the sum of potential energy and kinetic energy.

The potential energy of a spring is given by:

PE = (1/2)kx^2.

Substituting the provided values:

k = 100 N/m, x = 0.2 m,

we can calculate:

PE = (1/2) * 100 N/m * (0.2 m)^2 = 2 J.

The kinetic energy at the equilibrium position is zero, so the total mechanical energy of the system is equal to the potential energy. Thus, the total mechanical energy of the system is 2 J.

c) When the mass has a speed of half its maximum speed, its kinetic energy is also half of its maximum kinetic energy.

The maximum kinetic energy is equal to the total mechanical energy, which we calculated as 2 J.

At any displacement, the kinetic energy is given by:

KE = (1/2)mv^2,

where KE is the kinetic energy, m is the mass, and v is the velocity.

Setting KE equal to half of the maximum kinetic energy:

(1/2)mv^2 = (1/2) * 2 J,

We can solve for v:

v^2 = 4 J / (2.5 kg) = 1.6 m^2/s^2,

v = sqrt(1.6) m/s ≈ 1.26 m/s.

We can find the corresponding displacement by using the equation for the kinetic energy:

(1/2)(2.5 kg)(1.26 m/s)^2 = (1/2)kx^2.

Substituting the provided value for k = 100 N/m, we can solve for x:

(1/2)(2.5 kg)(1.26 m/s)^2 = (1/2)(100 N/m)(x^2),

1.58 J = 50 N/m(x^2),

0.0316 m = x^2,

x ≈ ±0.178 m.

Since the displacement cannot be negative, the displacement at which the mass has a speed of half its maximum speed is 0.178 m.

d) The period and frequency of the motion can be determined using the equation for the period of simple harmonic motion:

T = 2πsqrt(m/k),

where T is the period, m is the mass, and k is the spring constant.

Substituting the provided values:

m = 2.5 kg, k = 100 N/m,

we can calculate:

T = 2πsqrt(2.5 kg / 100 N/m) ≈ 0.5 s.

The frequency is the reciprocal of the period:

f = 1/T = 1/0.5 s = 2 Hz.

Therefore, the period of the motion is 0.5 s and the frequency is 2 Hz.