A 1.2-kg block sliding at 5.0 m/s on a frictionless surface runs into and sticks to a spring. The spring is compressed 0.10 m before stopping the block and starting its motion back in the opposite direction.

1. If vibrations start after the collision, determine their amplitude. (A=?)

2. Determine Spring constant (k=?)

3. Determine frequency of simple harmonic motion (f=?)

Answer: Amplitude A = 0.1414m, spring constant (k) = 1500N/m, frequency of harmonic motion (f) = 5.262Hz.

Explanation: to get the Amplitude of the motion, we have to consider the law of conservation of energy.

The kinetic energy of the spring equals the elastic potential energy stored in the spring.

Kinetic energy = 1/2mv²

Elastic potential energy = 1/2 KA².

Where m = mass of loaded spring=1.2kg

v = velocity of motion=5m/s

k = elastic constant = 1500N/m

A = Amplitude = unknown

1) Amplitude.

1/2mv² = 1/2 KA²

Thus we have that

mv² = KA²

1.2 * 5² = 1500 * A²

1.2 * 25 = 1500 * A²

30 = 1500 * A²

A² = 30/1500

A² = 0.02

A = (0.02) ^1/2

A = 0.1414m.

2) spring constant.

On hitting the spring, the mass does not accelerates any more, it now decelerates making the final velocity become an initial velocity and the acceleration turns to deceleration (-a).

We use Newton laws of motion to get the magnitude of deceleration.

v² = u² + 2ax

Where v = final velocity = 0 (because the mass is coming to rest), u = initial velocity = 5m/s, x = compression = distance covered when the mass hits the spring = 0.10m

But substituting parameters, we have

0² = 5² + 2 (-a) (0.1)

0 = 25 - 0.2a

0.2a = 25

a = 125m/s

We get the spring constant using hooke's law which needs force, thus

F = m*a

F = 1.2 * 125 = 150N.

From hooke's law

F = ke

Where F = force = 150N, k = elastic constant and e = extension = 0.1m.

150 = k (0.1)

k = 150 / 0.1

k = 1500N/m.

3) frequency

Formula for frequency of a simple harmonic motion is

f = 1/2π * √ k/m

f = 1/2 * 3.142 * √ (1500/1.2)

f = 1 / 6.284 * √1250

f = 1/6.284 * 35.355

f = 5.626Hz