A physics lab is demonstrating the principles of simple harmonic motion (SHM) by

Question

A physics lab is demonstrating 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 355.0 g from the spring, the student notices the spring is displaced 45.5 cm from its previous equilibrium With this information, calculate the spring constant. Number N/ m When the spring, with the attached 355.0-g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the period of oscillation, T, neglecting the mass of the spring itself Number In the final section of the lab, the student is asked to investigate the energy distribution of the spring system described above. The student pulls the mass down an additional 34.1 cm from the equilibrium point of 45.5 cm when the mass is stationary and allows the system to oscillate. Using the equilibrium point of 45.5 cm as the zero point for total potential energy, calculate the velocity and total potential energy for each displacement given and insert the correct answer using the "choices" column. (Scroll down to see the full table.)

Explanation / Answer

m = 355.0 gm = 0.355 Kg
x = 45.5 cm = 0.455 m

We know,
F = k*x
m*g = k*x
0.355*9.8 = k*0.455
k =  7.65 N/m

Time period is given by,
T = 2* * sqrt(m/k)
T = 2* * sqrt(0.355/7.65)
T = 1.35 s


When Displacement = 34.1 cm
Total Potential Energy = 1/2 * kx^2
P.E = 1/2 * 7.65 * 0.341^2 J
P.E = 0.445 J

Using Energy Conservation,
1/2 * mv^2 = 0.445
1/2 * 0.355 * v^2 = 0.445
v = 1.58 m/s

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