Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0, the block has velocity -4.00 m/s and displacement +0.200 m. Write an equation for the position as a function of time.

Answer 1

The position as a function of time for the block attached to a spring undergoing SHM can be written as x(t) = A cos(10πt + ϕ), where A is the amplitude and ϕ is the phase constant, calculated from the initial conditions.

Step-by-step explanation:

To find the equation for the position as a function of time for a 2.00-kg block attached to an ideal spring with a force constant of 300 N/m, we use simple harmonic motion (SHM) principles. Given at t = 0, the block has a velocity of -4.00 m/s and displacement of +0.200 m. The general solution for SHM is expressed as x(t) = A cos(ωt + ϕ). Where:

• ω (angular frequency) = √(k/m)
• A (amplitude) = √(x02 + (v02/ω2))
• ϕ (phase constant) is determined from initial conditions.

The angular frequency ω for our spring-mass system is √(300/2.00) = 10 π rad/s. With the initial conditions, we can calculate the amplitude A and the phase ϕ. Using the initial velocity, v0 = -4.00 m/s, we solve for ϕ.

The final equation of motion becomes x(t) = A cos(10πt + ϕ), with specific values for A and ϕ calculated from given conditions.