A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.200 m and the period is 3.35 s . a.) What is the acceleration when x = 0.160? b.) What is the speed of the block when x= 0.160 m ?

a = 0.563 m/s²

v = 0.225 m/s

Explanation:

The angular frequency is:

ω = 2π / T

ω = 2π / (3.35 s)

ω = 1.88 rad/s

For a spring mass system:

ω = √(k / m)

ω² = k / m

(a) Sum of forces on the mass:

∑F = ma

kx = ma

a = kx / m

a = xω²

When x = 0.160 m:

a = (0.160 m) (1.88 rad/s)²

a = 0.563 m/s²

(b) Energy is conserved.  Initial elastic energy = final elastic energy + kinetic energy.

EE₀ = EE + KE

½ kA² = ½ kx² + ½ mv²

½ (k/m)A² = ½ (k/m)x² + ½ v²

½ ω²A² = ½ ω²x² + ½ v²

ω²A² = ω²x² + v²

v² = ω² (A² − x²)

v = ω √(A² − x²)

v = (1.88 rad/s) √((0.200 m)² − (0.160 m)²)

v = 0.225 m/s

newton's first law of motion