tion 8 of 12 Sapling Learning A physics lab is demonstrating the principles of s

Question

tion 8 of 12 Sapling Learning A physics lab is demonstrating the principles of simple harmonic motion (sHM) by using a spring 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 27.5 cm from its previous equilibrium. With this information, calculate the spring constant. When the spring, with the attached 315.0-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 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 20.6 cm from the equilibrium point of 27.5 cm when the mass is stationary and allows the system to oscillate. Using the equilibrium point of 27.5 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.) Displacement (cm) Velocity (m/s) Total Potential Energy (J) Choices equilibrium 0.837 20.6 O Previous Check Answer Next Exit about us careers privacy policy terms of use contact lunteerwaiver.....docx

Explanation / Answer

(a)

Given that,

m = 315 g = 0.315 kg

x = 27.5 cm = 0.275 m

We know that,

F = kx = mg

k = mg / x = 0.315*9.8 / 0.275

k = 11.22 N/m

(b)

period of oscillation,

T = 2*pi*sqrt (m / k)

T = 2*3.14*sqrt(0.315 / 11.22)

T = 1.052 s

(c)

velocity v = (2pi / T)*sqrt (A^2 - x^2)

when x = 20.6 cm

v = 2pi / 1.052 * sqrt [(0.275)^2 - (0.206)^]

v = 1.087 m/s

Total energy = KE + PE = (1/2)*mv^2 + (1/2)*kx^2

TE = (1/2)*0.315*(1.087)^2 + (1/2)*11.22*(0.206)^2

TE = 0.424 J

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