A mass m = 50g is suspended from a spring with spring constant k = 8.0N/m. Deter

Question

A mass m = 50g is suspended from a spring with spring constant k = 8.0N/m. Determine the force acting on the mass when it is displaced 8cm from its equilibrium position. If the mass is then released, what is the period of oscillation? What is the frequency? What is expression for the uncertainty in the total energy E_tot of equation 3.5? Assume an uncertainty in k of delta k, uncertainty in m of delta m, uncertainty in x of delta x and an uncertainty in v of delta v. Assume g is perfectly known (delta g = 0). Refer to the section on the Uncertainty Calculations in the introduction, Section 0.2.2, for more details.

Explanation / Answer

weight of m = mg = (0.050)(9.81) = 0.4905 N {acts DOWN}

F = k(x1)

x1 = 0.4905/8.0 = 0.0613125 0.061 = 6.1 cm

{x1 = initial spring stretch due to weight of m is the equilibrium stretch position of spring-mass system}

8 cm additional stretch = 8 + 6.1 = 14.1 = 0.141 m max stretch of the spring

F = k(0.141) = (8.0)(0.141) = 1.128 1.13 N {total downward force reqd for max stretch}

When the spring is displaced 8 cm from its equilibrium position, the spring restoring force (acts vertically UPwards) on the mass, m, = 1.13 N. {acts UP}

The NET force acting on mass, m,

when spring is displaced 8 cm from its equilibrium position = 1.13 UP - 0.49 0.64 N {acts UP}

The period T is found by its formula of definition:

T = 2m/k

so T=0.496

and frequency = 1/T=2

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