A small block is attached to an ideal spring and is moving in SHM...

Question

A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the motion is 0.165 m. The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of the block?

Accepted Answer

Welcome back everybody. We are looking at a particle that is in simple harmonic motion. We are told that it has an amplitude of 25 centimeters or 250. m were told that it has a velocity at the equilibrium point of 0.0.5 meters per second or five centimeters a second. And we are asked to find what the maximum acceleration is going to be. Well, the acceleration of the particle at any point of harmonic motion is given by this formula of negative omega squared times its amplitude times the cosine of omega times time plus five. Now, here's the thing, this is going to be maximum when this cosine value is maximum or the absolute value of this Cosine values at maximum, which happens when this cosine value is equal to one, meaning that our maximum acceleration is just equal to omega squared a let's see which is equal to K over M times our amplitude. Now we don't know what the mass or the spring constant of our particle and system is, right. So we need to find out why what this ratio is. And the way we're going to do that by using the energy conservation principles of harmonic motion, which states that our kinetic energy or one half times mass times velocity squared plus our potential energy, which is one half times K, times X squared is equal to one half K times a squared. Now our equilibrium point or at X equals zero. This just says that one half times our mass times our velocity at our equilibrium squared is equal to one half times are spring constant times our amplitude squared, solving for the ratio of K over M. We get that it is our equilibrium velocity over our amplitude squared. Let's go ahead and plug in the values for that. This is equal to .05, divided by .25 squared, which gives us 0. Seconds to the native 2nd. Great! Now that we have that, let's go ahead and revisit our earlier formula that our acceleration at our maximum point is equal to this ratio that we found times our amplitude and plug in our values 0.04 seconds to the negative second times are amplitude of 0. m. Giving us a maximum acceleration of 0.1 m per second squared corresponding to answer choice. A Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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