Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

A 297 g block is connected to a light spring with spring constant 4.34 N/m, and displaced 7.45 cm from equilibrium. It is then released and allowed to oscillate in simple harmonic motion. What is the maximum acceleration of the block

Sagot :

Answer:

x = A sin ω t       describes the displacement of the particle

v = A ω cos ω t

a = -A ω^2 sin ω t        

a (max) = -A ω^2     is the max acceleration (- can be ignored here)

ω = (K/ m)^1/2        for SHM

F = - K x^2     restoring force of spring

K = 4.34 / .0745^2 = 782 N / m

ω = (782 / .297)^1/2 = 51.3 / sec

a (max) = .0745 * 782 / .297 = 196 m / s^2

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.