(II) A 0.350-kg mass oscillates on the end of a thin vertical ...

Question

(II) A 0.350-kg mass oscillates on the end of a thin vertical spring (spring constant 265 N/m) with an amplitude of 28.0 cm. The mass passes through the equilibrium point (y = 0) with positive velocity at t = 0 .

(b) At what times will the length of the spring be longest and shortest?

(b) At what times will the length of the spring be longest and shortest?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem 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. At what time will a 0.262 kg ball swing on a slender vertical spring spring constant, 306 newtons per meter achieve its longest and shortest length if it starts oscillating with an amplitude of 29.0 centimeters and crosses the equilibrium point Y equals zero upwards at T equals zero. So it appears that the entire problem itself is all important information, but we're asked to solve for two separate answers. Our first answer is we're asked to figure out what time will it take for a 0.262 kg bob to swing on a cylindrical on a sl slender vertical spring to achieve its longest and shortest lengths if it starts oscillating with an amplitude of 29.0 centimeters and crosses the equilibrium point Y equals zero and upwards at T equals zero seconds. So we're trying to find the maximum time and the minimum time So looking at our multiple choice answers, we can see we have an answer for T max, which is T subscript max first and then our second answer will be for T subscript min or T minimum. And for both T max and T minimum at the very end, it's for the conditions, N equals zero, comma one comma two dot dot dot And this is the same for all of our multiple choice answers for both T max and for T min. So let's read them off to see what our final answer pair might be. So A is 1.60 multiplied by 10 to the power of negative two seconds plus N multiplied by 0.184 seconds and 0.37 multiplied by 10 to the power of negative one seconds plus N multiplied by 1.148 seconds. B is 2.00 multiplied by 10 to the power of negative two seconds plus N multiplied by 0.104 seconds and 0.36 multiplied by 10 to the power of negative one seconds plus N multiplied by 0.111 seconds. C is 4.60 multiplied by 10 to the power negative two seconds plus N multiplied by 0.184 seconds and 1.38 multiplied by 10 to the power negative one seconds plus N multiplied by 0.184 seconds. And finally, for D 6.11 multiplied by 10 to the power of negative two seconds plus N multiplied by 0.124 seconds and 3.83 multiplied by 10 to the power of negative one seconds plus N multiplied by 0.190 seconds. OK. So first off, we need to assume that the downwards direction is the positive direction of motion. Now, we need to recall and write down the equation for the frequency of oscillation which states that F is equal to one divided by two multiplied by pi multiplied by the square root of K divided by M. And we can also write out that F is also equal to two multiplied by pi multiplied by F which is equal to the square root of K which is the spring constant divided by the mass. And once again, F is the frequency. So we could rearrange this equation to isolate and sulfur omega. And we can also note that Omega which is the angular frequency is equal to or I should say the, I said angular frequency. But I meant to say that Omega is the angular velocity. So Omega the angular velocity which is equation one is equal to the spring constant divided by the mass. Awesome. And once again, F is the frequency of oscillation and M is the mass of the object. So now at this point, we need to plug in all of our known variables into equation one to solve for the numerical value of omega. So we're given the spring constant as 306 newtons per meter divided by our mass, which is given to us as 0.262 kg. And when we plug that into our calculator, we will get 34.175 radiance per second when we round to three decimal places. So now at this point, we need to recall and use the equation for the period of oscillation which states that T the period is equal to two multiplied by pi divided by Omega, which is also equal to two multiplied by pi divided by our value of Omega, which we found to be 34 multiply. So 34.175. So 34.175 don't forget. Its units are radians per second. So when we plug that into our calculator and we round to three decimal places, we will get 0.184 seconds. So now at this point, we need to note that the spring will have its maximum extensions at the following times. So at T max, it will be equal to T divided by four plus N multiply by T which is equal to 4.60 multiplied by 10 to the power of negative two seconds plus N multiplied by 0.184 seconds. And this is for the conditions N equals zero, comma one comma two comma dot dot dot And similarly, the spring will have its minimum extensions at the following times. So at T min is equal to three multiplied by T divided by four plus N multiplied by T which is equal to 1.38 multiplied by 10 to the power of negative one seconds plus N multiplied by 0.184 seconds which is equal to. So N is equal to zero, comma one comma two dot dot dot And that's it we've solved for this problem. Hooray, we did it. So now we just need to look at our multiple choice answers to see which answer is our final answer. Choice that matches the answers that we just found together, which it appears that multiple choice answer letter C has to be our final and correct answer. So T max is equal to 4.60 multiplied by 10 to the power of negative two seconds plus N multiplied by 0.184 seconds and then N equals zero comma one comma two comma dot dot dot And T minimum is equal to 1.38 multiplied by 10 to the power negative one seconds plus N multiplied by 0.184 seconds. Comma N equals zero, comma one comma two comma dot dot dot And that's it we've solved for this problem. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

(II) A 0.350-kg mass oscillates on the end of a thin vertical spring (spring constant 265 N/m) with an amplitude of 28.0 cm. The mass passes through the equilibrium point (y = 0) with positive velocity at t = 0 .

(b) At what times will the length of the spring be longest and shortest?

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Struggling with Physics?

(II) A 0.350-kg mass oscillates on the end of a thin vertical spring (spring constant 265 N/m) with an amplitude of 28.0 cm. The mass passes through the equilibrium point (y = 0) with positive velocity at t = 0 .(b) At what times will the length of the spring be longest and shortest?

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(II) A 0.350-kg mass oscillates on the end of a thin vertical spring (spring constant 265 N/m) with an amplitude of 28.0 cm. The mass passes through the equilibrium point (y = 0) with positive velocity at t = 0 .

(b) At what times will the length of the spring be longest and shortest?