Simple harmonic motion of a mass-on-a-vertical-spring is being demonstrated to a

Question

Simple harmonic motion of a mass-on-a-vertical-spring is being demonstrated to a physics class. Studen Map t are asked to find the spring constant, k. After suspending a mass of 315.0 g from the spring, one student notices the end of the spring is displaced 25.5 cm from its previous equilibrium. With this information, calculate the spring constant. Number 12.105 N/ m When the spring, with the attached 315.0-g mass, is displaced from its new equilibrium position, it undergoes Calculate the period of oscillation, T, neglecting the mass of the spring itself Number In the final section of the class, a student is asked to investigate the energy distribution of the spring system described above. The student pulls the mass down an additional 19.1 cm from the equilibrium point of 25.5 cm (where the mass had hung stationary), and releases the mass from rest, causing the system to oscillate, Using the equilibrium point of 25.5 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) from equilibrium Velocity (m/s) Total Potential Energy (J)Choices 0 0.119 1.18 3940 0.806 0.221 19.1 14.0

Explanation / Answer

(1)

Here,

kx = mg

spring constant k = mg / x

k = 0.315*9.8 / 0.255

k = 12.10 N.m

(2)

Period of oscillation,

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

T = 2*pi*sqrt (0.315 / 12.10)

T = 1.013 s

(3)

(a)

At displacement from equilibrium = 19 .1 cm

when amplitude is 19.1 cm KE is minimum and PE is maximum.

PE = (1/2)kA^2 = (1/2)*12.10*(0.191)^2

PE = 0.221 J

KE = 0 so, v = 0

(b)

at x = 14 cm

KE = (1/2)k(A^2 - x^2) = (1/2)mv^2

12.105*[(0.191)^2 - (0.14)^2] = 0.315 * v^2

v = 0.806 m/s

PE = (1/2)*kx^2

PE = (1/2)*12.105*(0.14)^2

PE = 0.119 J

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