A small 20-kg block A is attached to the rod B C of...

A small $20-\mathrm{kg}$ block $A$ is attached to the rod $B C$ of negligible mass that is supported at $B$ by a pin and bracket and at $C$ by a spring of constant $k=2 \mathrm{kN} / \mathrm{m}$. The system can move in a vertical plane and is in equilibrium when the rod is horizontal. The rod is acted upon at $C$ by a periodic force $P$ of magnitude $P=P_{m} \sin \omega_{f} t,$ where $P_{m}=6 \mathrm{N}$. Knowing that $b=200 \mathrm{mm}$, determine the range of values of $\omega_{f}$ for which the amplitude of vibration of block $A$ exceeds $3.5 \mathrm{mm} .$

Key Concepts

Dynamic Amplification Factor

The dynamic amplification factor quantifies how much larger the amplitude of a forced vibration is compared to the static displacement under the same load. It is crucial in determining the conditions under which the vibration amplitude exceeds a given threshold. By understanding this factor, one can predict the frequency range over which the system experiences large oscillations compared to its static deflection.

Natural Frequency and Resonance

The natural frequency is the rate at which a system oscillates when it is disturbed from its equilibrium position and allowed to vibrate freely. Resonance occurs when the frequency of the external force matches the natural frequency of the system, leading to maximum amplitude response. Knowledge of these concepts is essential to analyze conditions where small periodic forces produce large oscillatory responses.

Forced Vibrations

Forced vibrations refer to the motion induced in a system when an external time-varying force acts upon it. In mechanical systems, when a periodic force is applied, the response of the system is determined by the characteristics of the system as well as the frequency of the forcing. This concept is central to understanding how systems behave under dynamic loading and how the amplitude of motion depends on the forcing frequency.

Frequency Response of Undamped Systems

The frequency response of a system describes how the amplitude of its steady?state response varies with the frequency of the applied force. In undamped or lightly damped systems, this response is particularly sensitive near resonance, and even a small force can produce significant vibrations. This reasoning underlies the analysis of amplitude thresholds in response to a periodic load.

Transcript

00:01 Hi, everybody.

00:02 So determine the range of values of your frequency for which the amplitude, vibration of the block exceeds 3 .5, okay? and let me just draw this.

00:16 And what you're going to do exactly what it said.

00:18 You're just going to find the frequency, okay? i mean, there's no really better explanation than what the problem actually says.

00:28 And you're using, you know, you're mechanics, okay? if you don't know what this book is about, the book's about mechanical physics, okay? so you should know that.

00:41 And there's not much to say else besides we're going to just do follow along here with me because otherwise there's no point in sitting there detailing what i'm doing.

00:58 Because i am telling you what i'm doing every step of the way.

01:02 Okay.

01:06 Here we got the, i'm just kind of copying.

01:12 Oops.

01:20 There we go.

01:28 Okay.

01:29 I'm just copying the figure, by the way.

01:31 If you're curious, what i'm doing, and this, what i'm doing is just copying the figure.

01:37 I'm not doing anything special here.

01:39 Okay.

01:40 So in figure one, the angle is placed in that given and rod b .c.

01:46 Is the angle.

01:47 And the angular acceleration of the block of a is the double and triple of the theta.

01:56 So what we have to do is take the moment about b.

02:01 Okay.

02:02 So take the moment about b.

02:05 So right here, you see this b here.

02:07 You have to take the moment about it.

02:09 So looks like the force is going up.

02:15 So you're just going to have to do the right hand rule.

02:19 Okay.

02:22 And let's see.

02:23 Let's write this down for everybody.

02:26 So we have the math.

02:27 Mass of b equals so we have to do clockwise okay so we're going to do the right hand alone so we're going to do a little handy motion and you're going to be it's going to go clockwise okay times b remember that this right here is equal to a block okay and but i'm just going to delete it because i want you be able to see so you see his nb okay and it goes about this be here you see here how it's making a little triangle okay and i want to put clockwise so you guys know what direction it is plus and we have spring constant times the l times the theta which is this guy over here times cross -body product can guess which what is of this l here again just making a little um triangle clockwise.

03:38 Okay.

03:41 And now we have the p here and the l.

03:48 Okay.

03:50 And we're going to put cw clockwise.

03:53 Okay.

03:55 So if we do the cross -body product for each one, we get m, b squared, theta, beta, plus the k, l squared, beta, minus p .l.

04:09 L.

04:09 Okay.

04:11 So, at equilibrium, the sum of the moment about b is zero, thus equates mb is going to be zero of when it's at equilibrium...