Question:

A small block is attached to a horizontal spring with spring constant k. The block-spring system is placed on a frictionless surface. When the block is pulled back and released, it undergoes periodic motion.

a) Explain what is meant by periodic motion.

b) Derive the equation for the period (T) of the block's oscillation in terms of the mass of the block (m) and the spring constant (k).

c) A block with mass 0.5 kg is attached to a spring with a spring constant of 20 N/m. If the block is pulled back by 10 cm and released, calculate the period of its oscillation.

Answer:

a) Periodic motion refers to the motion of an object that repeats itself after a fixed interval of time. In other words, the motion is cyclical and follows a regular pattern.

b) To derive the equation for the period of the block's oscillation, we can start by considering Hooke's Law for a spring:

The force exerted by the 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.

We also know that the force acting on the block is given by Newton's Second Law of Motion:

F = ma

where m is the mass of the block and a is the acceleration.

By equating the two expressions for force, we have:

-kx = ma

Rearranging the equation, we get:

a = -(k/m)x

The acceleration (a) is related to the displacement (x) by the equation:

a = d²x/dt²

where dx/dt represents the velocity (v) of the block.

Substituting this equation into the previous one, we have:

d²x/dt² = -(k/m)x

This is a second-order linear differential equation, and its solution can be expressed as:

x(t) = A * cos(ωt + φ)

where A is the amplitude of motion, ω is the angular frequency, t is time, and φ is the phase constant.

The period (T) of the motion is the time taken for one complete cycle. It can be determined by noting that the argument of the cosine function in the equation for x(t) must increase by 2π for one full cycle. Therefore, we have:

ωT = 2π

Solving for T, we get:

T = 2π/ω

To find ω, we can rewrite the equation as:

ω = √(k/m)

Substituting ω back into the equation for T, we finally have:

T = 2π√(m/k) [Equation 1]

c) Given: m = 0.5 kg, k = 20 N/m

Using Equation 1, we can calculate the period (T):

T = 2π√(m/k) = 2π√(0.5/20) = 2π√(0.025) ≈ 6.28 * 0.158 ≈ 0.99 s

Therefore, the period of oscillation for a block with mass 0.5 kg and a spring constant of 20 N/m is approximately 0.99 seconds.