RaySix

Post

Oscillations of a Mass-Spring System

Created by @nathanedwards

at October 31st 2023, 4:18:20 pm.

AP_Physics_1-Questions

#displacement

#spring

#oscillation

Question:

A spring with a spring constant of 200 N/m is suspended vertically from a fixed point. A block of mass 2 kg is attached to the free end of the spring. The block is pulled downwards from its equilibrium position and then released.

a) Calculate the displacement of the block from its equilibrium position when it comes to rest for the first time.

b) Determine the maximum speed of the block while oscillating.

c) If the spring is compressed such that the block is displaced 0.3 m from the equilibrium position and released, calculate the time it takes for the block to complete one full oscillation.

Acceleration due to gravity, g = 9.8 m/s².

Answer:

a) To find the displacement of the block when it comes to rest for the first time, we need to consider the equilibrium position and the net force acting on the block.

According to Hooke's Law, the force exerted by a spring is proportional to the displacement from the equilibrium position. Mathematically, it can be represented as:

F = -kx

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

When the block comes to rest, the net force acting on it is zero. At equilibrium, the gravitational force mg acting downwards is balanced by the force exerted by the spring kx acting upwards. Hence:

mg = kx

Rearranging the equation, we can determine the displacement:

x = \frac{mg}{k}

Substituting the values:

x = \frac{(2 \, \text{kg})(9.8 \, \text{m/s²})}{200 \, \text{N/m}} = \frac{19.6}{200} \, \text{m} = 0.098 \, \text{m}

Therefore, the displacement of the block when it comes to rest for the first time is 0.098 m.

b) The maximum speed of the block while oscillating can be determined using the principle of conservation of mechanical energy. At the maximum displacement from the equilibrium position, all the potential energy stored in the spring is converted into kinetic energy.

The potential energy stored in a spring is given by the equation:

PE = \frac{1}{2} kx^2

The kinetic energy of the block is given by:

KE = \frac{1}{2} mv^2

At the maximum displacement, the potential energy is maximum and kinetic energy is zero. Therefore:

PE_{\text{max}} = \frac{1}{2} mv_{\text{max}}^2

\frac{1}{2} kx_{\text{max}}^2 = \frac{1}{2} mv_{\text{max}}^2

Rearranging the equation to solve for maximum speed:

v_{\text{max}} = \sqrt{\frac{k}{m}}x_{\text{max}}

Substituting the known values:

v_{\text{max}} = \sqrt{\frac{200 \, \text{N/m}}{2 \, \text{kg}}} \cdot 0.098 \, \text{m} = \sqrt{98} = 9.90 \, \text{m/s}

Therefore, the maximum speed of the block while oscillating is 9.90 m/s.

c) To calculate the time it takes for the block to complete one full oscillation, we can use the period formula for a mass-spring system:

T = 2\pi\sqrt{\frac{m}{k}}

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

Substituting the given values:

T = 2\pi\sqrt{\frac{2 \, \text{kg}}{200 \, \text{N/m}}} = 2\pi\sqrt{\frac{1}{100}} = 2\pi\frac{1}{10} = \frac{\pi}{5} \, \text{s}

Therefore, it takes the block π/5 seconds (approximately 0.628 s) to complete one full oscillation.