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

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 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when x = 0.160 m?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice prom together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A mechanical oscillator consists of a block of mass M and a spring of force constant K placed on a smooth horizontal table. The block undergoes simple harmonic motion with an amplitude 30 centimeters and a frequency 0.5 Hertz at the position X equals 20 centimeters. What are I, the speed V and I I the acceleration of the object A X. So that's our angle is we're asked to solve for two separate answers. So we're asked to solve for what the speed V is and what the acceleration of the object A X is when the position is at X equals 20 centimeters. So that'll be our final answer that we're ultimately trying to solve. For. Awesome. We're also given some multiple choice answers. Let us note that for all of our speed values, they're all gonna be in meters per second units. And for all of our acceleration values, which is for part I will be in units of meters per second squared. So all of our answers for part I are in units of meters per second. And all of our answers for part I, I are in meters per second squared. So let's read them off to see what our final answer pair might be. A is 0.01 and positive 0.31. B is 0.09 and positive 0.63 C is 0.17 and negative 1.25. And finally, for D it's 0.70 and negative 1.97. Awesome. So first off, let us recall and apply the principle of conservation of mechanical energy which states that em equals K plus U. So this is the mechanical energy equation. We can also write one half multiplied by M multiplied by VX squared. So VX square plus one half multiplied by lower case K multiplied by X squared is equal to one half multiplied by lower case K multiplied by capital A square. So now we need to consider the velocity at position X. And we could write that as VX is equal to plus or minus the square root of lowercase K divided by lowercase M multiplied by the square root of capital A squared minus lowercase X squared. So and the period of the simple harmonic oscillator. So to say it once again, so now we're writing the equation for the period of the simple harmonic oscillator. So the period equation could be written as and once again, this is for the simple harmonic motion of this oscillator to be specific is equal to two multiplied by pi multiplied by the square root of lower case M divided by lowercase K. Thus, the velocity at position X can be expressed as a function of the period T or the frequency F as VX equals plus or minus two multiplied by pi divided by T multiplied by the square root of capital A squared minus X squared, which is lowercase X equals plus or minus two multiplied by pi multiplied by F which F is the frequency multiplied by the square root of A squared which is capital A minus X squared. And it's lower case X. So now the speed of the block could be written as V equals the absolute value of plus or minus two multiplied by pi divided by T multiplied by the square root of capital A squared minus lowercase X squared, which we can write as V equals two pi or two multiplied by pi multiplied by F multiplied by the square root of capital A squared minus lowercase X squared. Awesome. So moving right along, we could write that V equals two multiplied by pi multiplied by 0.5 Hertz is we're putting in our known values and our frequency is given to us in the prom itself as 0.5 Hertz multiplied by the square root of, and our value for A is 0.3 meters because we're told that the amplitude is capital A in this case denotes the amplitude and we're given it as 30 centimeters, but we have to convert centimeters to meters to keep all of our units consistent. And don't forget that this is squared minus or X value, which we're given it given that value as X equals 20 centimeters. So writing it in units of meters, it's 0.2 m. And don't forget that that value is squared two. So when we plug that into our calculator and round two decimal places, we will find that it's equal to 0.70 m per second. And that is our final answer. For part I hooray, we did it and this is our speed of the block. Awesome. So now to find the acceleration of the particle at X equals 0.2 m is as follows. So AX is equal to negative K divided by M multiplied by X which is equal to negative two multiplied by pi divided by capital T all squared multiplied by X which is equal to negative two, multiplied by pi multiplied by F all squared multiplied by X. So, Ax is equal to negative two multiplied by pi multiplied by 0.5. Because 0.5 Hertz is our frequency value all squared multiplied by our value for X which is given to us as 20 centimeters which is 0.2 m. So when we plug that into our calculator and round two decimal places, we will get negative 1.97 m per second squared. And that is our final answer. For part I I hooray, we did it so the block's acceleration and so this is the block's acceleration and it is directed towards the negative X axis and will have a magnitude of negative 1.97 m per second squared. And this corresponds with multiple choice answer letter D which states that I is V equals 0.70 m per second and I I is A X equals negative 1.97 m per second squared. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

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 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when x = 0.160 m?

Key Concepts

Simple Harmonic Motion (SHM)

Amplitude and Period

Kinematics of SHM

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