A physics lab is demonstrating the principles of simple harmonic motion (SHM) by using a spring affixed to a horizontal

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899604
Joined: Mon Aug 02, 2021 8:13 am

A physics lab is demonstrating the principles of simple harmonic motion (SHM) by using a spring affixed to a horizontal

Post by answerhappygod »

A Physics Lab Is Demonstrating The Principles Of Simple Harmonic Motion Shm By Using A Spring Affixed To A Horizontal 1
A Physics Lab Is Demonstrating The Principles Of Simple Harmonic Motion Shm By Using A Spring Affixed To A Horizontal 1 (272.64 KiB) Viewed 24 times
A Physics Lab Is Demonstrating The Principles Of Simple Harmonic Motion Shm By Using A Spring Affixed To A Horizontal 2
A Physics Lab Is Demonstrating The Principles Of Simple Harmonic Motion Shm By Using A Spring Affixed To A Horizontal 2 (222.35 KiB) Viewed 24 times
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 255.0 g from the spring, the student notices the spring is displaced 23.5 cm from its previous equilibrium. With this information, calculate the spring constant. spring constant: N/m When the spring, with the attached 255.0 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. T = S In the final section of the lab, the student is asked to investigate the energy distribution of the spring system described previously. The student pulls the mass down an additional 17.6 cm from the equilibrium point of 23.5 cm when the mass is stationary and allows the system to oscillate. Using the equilibrium point of 23.5 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. Displacement (cm) from equilibrium Velocity (m/s) Total potential energy (J) *********** ********************
In the final section of the lab, the student is asked to investigate the energy distribution of the spring system described previously. The student pulls the mass down an additional 17.6 cm from the equilibrium point of 23.5 cm when the mass is stationary and allows the system to oscillate. Using the equilibrium point of 23.5 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. Displacement (cm) from equilibrium Velocity (m/s) Total potential energy (J) 17.6 12.9 Answer Bank 0 0.165 0.294 1.14 0.77 0.0889
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply