A physics lab demonstrates the principles of simple harmonic motion (SHM) by usi

Question

A physics lab demonstrates the principles of simple harmonic motion (SHM) by using a spring affixed to a horizontal support. The student is asked to find the spring constant, k. After suspending a mass of 315.0 g from the spring, the student notices the spring is displaced by 27.5 cm from equilibrium. With this information, calculate the spring constant.

= ___________ N/m

The student realizes that the spring demonstrates SHM with the attached mass of 315.0 g. The student is next asked to calculate the period of oscillation, T, neglecting the mass of the spring.

= ___________ s

For the final section of the lab, the student is asked to investigate the energy distribution of the spring system described above. Calculate the velocity and potential energy for each displacement given and insert the correct answer using the "choices" column.

Explanation / Answer

kx = mg
k = mg/x = 0.315*9.8/0.275 = 11.23N/m

Time Period T = 2(m/k) = 1.053s

Displacement = 27.5cm

Here the velocity is 0

Potential energy = 0.5kx^2 = 0.5*11.23*0.275^2 = 0.425

Displacement = 15.5cm

velocity = *(A^2 -x^2) = 1.36m/s

A = 0.275m x = 0.155m

= (k/m) = 5.97

velocity = 5.97*(0.275^2 -0.155^2) = 1.36m/s

Total energy = 0.5kx^2 = 0.5*11.23*0.275^2 = 0.424J

Kinetic energy = 0.5mv^2 = 0.5*0.315*1.36^2 = 0.291J

Potential energy = 0.424 - 0.291 = 0.135J

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