Question

The maximum speed of a 4.1 kg mass attached to spring is 0.78 m/s and the maximum force exerted on the mass is 13 N.

Answer 1

Start with the conservation of energy. The spring potential energy and the mass' kinetic energy will fluctuate over time, but their sum will stay constant. The maximum spring potential energy equals the maximum kinetic energy.

0.5mv² = 0.5kx²

m is the mass, v is the maximum velocity, k is the spring constant, and x is the maximum displacement along the spring.

Given values:

m = 4.1kg

v = 0.78m/s

Calculate the maximum kinetic energy.

Max KE = 0.5mv² = 1.247J

Set this equal to the maximum spring potential energy.

Max spring PE = 0.5kx² = 1.247J

x² = 2.494/k

The spring force is F = kx

Max F = kx = 13N

x = 13/k

x² = 169/k²

Set both values of x² equal to each other and solve for k the spring constant:

2.494/k = 169/k²

2.494k = 169

k = 67.8N/m

Use k to find x:

Max F = kx = 13N

67.8x = 13

x = 0.192m

The frequency of the system is given by:

f = (1/(2π))√(k/m)

f is the frequency, k is the spring constant, m is the mass.

f = (1/(2π))√(67.8/4.1)

f = 0.647Hz