Respuesta :
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
