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A diving board oscillates with simple harmonic motion of frequency 2.8 cycles per second. What is the maximum amplitude with which the end of the board can oscillat...

Question

A diving board oscillates with simple harmonic motion of frequency 2.8 cycles per second. What is the maximum amplitude with which the end of the board can oscillate in order that a pebble placed there (Fig. 11$-$56) does not lose contact with the board during the oscillation?

A diving board oscillates with simple harmonic motion of frequency 2.8 cycles per second. What is the maximum amplitude with which the end of the board can oscillate in order that a pebble placed there (Fig. 11$-$56) does not lose contact with the board during the oscillation?



Answers

A diving board oscillates with simple harmonic motion of frequency 2.8 cycles per second. What is the maximum amplitude with which the end of the board can oscillate in order that a pebble placed there (Fig. 11$-$56) does not lose contact with the board during the oscillation?

So the frequency off our simple pendulum is, if equals to one over. To buy Times Square, root off G over l, and we get that from equation 11 point 11 beat. So the pendulum is basically accelerating. Vertical leverage is equivalent to increasing or decreasing the acceleration due to gravity by the acceleration of the pendulum. So the new frequency will be won over to buy Times Square root of D plus A. That's the increase in ax elation divided by N so we know that this is one over to buy square root off D plus 0.35 g divided by L or one over to buy square root of 1.35 g over. And so now we can take the constant or we can take 1.35 outside. So we have one by Dubai Times Square root of 1.35 square. Load off G over l. So that's equal. So squared off 1.35 times at where f is the frequency defined at the beginning of the problem. So that means if we evaluate this, we have a little bit of space here. So let's write down that. So this is 1.16 So it's one point 1/6 off the initial frequency. Now. For part B, the new frequency would be again one by two. Pi squared off G plus a buy. And where now, instead of plus, we have a minor science of Jean minus 0.35 g. And then we do. I did by l. So we can carry for the same calculation as before. And we see that we have 0.65 times, one by two bys credit of G over l. What is square root off? 0.65 f our 0.81 So it's 0.81 Ah, off the initial different frequency. All right, Thanks for watching. I'll see you next time.

For this problem on the topic of oscillators, we are told that a simple pendulum is oscillating with a frequency of F. We want to calculate the new frequency if the entire pendulum accelerates with half the acceleration due to gravity firstly upward and then downward. Now the frequency of a simple pendulum wino is given by F. This is one over two pi times the square root of G over L. So the pendulum is accelerating vertically, which is equivalent to increasing or decreasing the acceleration due to gravity by the acceleration of the pendulum. So let's first look at the scenario in which the pendulum is accelerating upward, so we'll call the new frequency F. New, This frequency is one over two pi. I'm the square root of G plus A. Of our, So this is one over two pi times the square root Of 1.5 G. Over our, or simply The square root of 1.5 times one over two pi times the square root of G over L. So the term in brackets is simply F. So the new frequency is the square root of 1.5 times the old frequency F. Which is simply 1.22 times the original frequency f. So the new frequency is greater than the old frequency. Now, what happens if the pendulum is accelerating downward? So some little above the new frequency F. New is one over two pi times the square root of G plus A. Over L. Hey in this case is going to change, that's one of the two pi times The Square Root of G -0.5 G. Which gives us 0.5G, divided by L. And again we can write this as the square root of a half Skirt of 0.5 Times 1/2 Pi. And the square root of G. Of L. Of the time in brackets again, is the original frequency F. This becomes a square root of a half times the original frequency F. And so this becomes 0.71 times F, which is now less than the original frequency.

So here the maximum downwards acceleration has to be equal to gravity because the board, of course, cannot move faster than the pebble. If the pebble on the board do not want to lose contact with one another, so the maximum acceleration would be equal, rather would be equal to four pies square times the frequency squared times the amplitude and this has to be less than G. Therefore, the amplitude must be less than or equal to this acceleration due to gravity divided by four pi squared times, the frequency squared. And so uh, we can say that the amplitude must be less than or equal to on 9.8 meters per second squared, divided by four pi squared times the frequency of 2.5 hertz squared and we find that the amplitude cannot be has to be less than or equal to 0.397 meters. So this would be Earth answer that is tthe e end of the solution. Thank you for watching

So here we know the frequency is equal to their cyclical of the period. This would be equal to one over one, over to pie multiplied by the square root of this spring. Constant divided by the mass solving for the spring constant In terms of the frequency, this would be equal to four pi squared, frequency squared, multiplied by mass. We know that the energy is gonna be equal to the potential Energy stored in the spring would be equal to 1/2 times the spring. Constant times the amplitude squared. This would be equal to 1/2 multiplied by four pi squared frequency squared times the mass multiplied by the amplitude squared and we can solve. The energy would then be equal to two pi squared, multiplied by the frequency of 2.8 hertz quantity squared, multiplied by the mass of 1.6 kilograms multiplied by the amplitude of 0.73 meters quantity squared me. Potential energy stored in the spring would then be equal to 1.3 jewels. This is our final answer. That is the end of the solution. Thank you for watching


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