Question

A small block is attached to an ideal spring and is moving in shm on a horizontal, frictionless surface. when the amplitude of the motion is 0.090 m , it takes the block 2.47 s to travel from x = 0.090 m to x = -0.090 m .

Answer 1

The equations of motion for a block attached to an ideal spring undergoing Simple Harmonic Motion (SHM) can be expressed as x(t) = Acos(ωt + φ), v(t) = -Aωsin(ωt + φ), and a(t) = -Aω²cos(ωt + φ). In this specific case, the block takes 2.47 seconds to travel from x=0.09 m to x=-0.09 m, which corresponds to one complete oscillation from the maximum positive displacement to the maximum negative displacement.

Step-by-step explanation:

The equations of motion for a block attached to an ideal spring undergoing Simple Harmonic Motion (SHM) can be expressed as:

x(t) = Acos(ωt + φ)

v(t) = -Aωsin(ωt + φ)

a(t) = -Aω²cos(ωt + φ)

Where:

• x(t) is the displacement of the block from the equilibrium position at time t
• A is the amplitude of the motion
• ω is the angular frequency
• φ is the phase angle

In this specific case, the block takes 2.47 seconds to travel from x=0.09 m to x=-0.09 m, which corresponds to one complete oscillation from the maximum positive displacement to the maximum negative displacement. Therefore, we can use the equation T = 2π/ω to find the period T, and then use it to find the angular frequency ω.