(1 point) An oscillating mass at the end of a spring is at a distance y from its equilibrium position given by y=????sin(????m‾‾‾√⋅????). The constant ???? measures the stiffness of the spring. Find the first positive time time ????1 at which the mass is farthest from its equilibrium position: ????1= Find the first positive time ????2 at which the mass is moving fastest: ????2= Find the first positive time ????3 at which the mass has the largest acceleration (in magnitude): ????3= What is the period, T, of the oscillation? T= Find ????T/????m: ????T????m= (What does the sign of ????T/????m tell you?)

Answer: See descriptionStep-by-step explanation:the position is given by [tex]y = Asin(\sqrt{\frac{k}{m} }*t)[/tex]a) the first time at which the mass is the farthest from its equilibrium position will be determined by the first value in which our y function is maximum; as it is a sine function the maximum values are [1,-1], we are taking 1 as the maximum and after replacing the sine function with 1 we have [tex]y = A[/tex].Replace y with A and solve :[tex]A = Asin(\sqrt{\frac{k}{m} }*t) => 1 = sin(\sqrt{\frac{k}{m} }*t) => sin^{-1}(1) = \sqrt{\frac{k}{m} }*t\\ \frac{\pi }{2} = \sqrt{\frac{k}{m} }*t => t = \frac{\pi }{2} / \sqrt{\frac{k}{m} }[/tex]b) In a simple harmonic motion the speed will be maximum whenever the position is 0, therefore we replace in our y equation[tex]0 = Asin(\sqrt{\frac{k}{m} }*t) => 0 = sin(\sqrt{\frac{k}{m} }*t)\\ sin^{-1}(0) = 0=> t = 0[/tex]) then 0 is the first time when the speed is maximum. You can think about it as a pendulum: whenever the ball reaches 0 speed it starts to come back untill its velocity is maximum in the equilibrium position, then it starts ascending again and losing energy.c) we derive the position equation as follows to obtain velocity:[tex]\frac{dy}{dt} = A*\frac{d(sin(\sqrt{\frac{k}{m} }*t) )}{dt}* \frac{d(\sqrt{k/m}*t)}{dt}  \\=v(t)= A\sqrt{k/m}*cos(\sqrt{k/m}*t)[/tex]then we derive again to obtain acceleration:[tex]\frac{dv}{dt} = A*\sqrt{k/m}*\frac{d(cos(\sqrt{\frac{k}{m} }*t) )}{dt}* \frac{d(\sqrt{k/m}*t)}{dt}  \\=a(t)= -A*k/m*sin(\sqrt{k/m}*t)[/tex]We follow the same logic as before, the acceleration is maximum whenever the sine function reaches [1 or -1]. In order to remove the minus sign we have in the equation we need to choose -1:[tex]a_{max}= -A*k/m*(-1) = A*k/m[/tex]then[tex]A*k/m = -A*k/m*sin(\sqrt{k/m}*t)\\ -1 = sin(\sqrt{k/m}*t)\\ sin^{-1} = \sqrt{k/m}*t = 3\pi /2[/tex]then solve t:[tex]t = 3\pi /2 * \sqrt{m/k}[/tex]d) the period is the time that the mass needs to go back and forth, and it can be obtained from the first equation. the logic is simple, the sine function is maximum every t = 1/2 (4 π n + π). With n = 0, 1, 2 we get pi/2, 5pi/2, and 9pi/2.Now we take the arguments of the sine function and we pick two of these values and then subtract them to get the period T:[tex]t*\sqrt{\frac{k}{m} }= \frac{\pi }{2}\frac{5\pi }{2}, \frac{9\pi }{2}\\ => t1 = \sqrt{\frac{m}{k} }*\frac{\pi }{2}\\ t2 = \sqrt{\frac{m}{k} }*\frac{5\pi }{2}\\ => T= t2-t1 = 2\pi*\sqrt{\frac{m}{k} }[/tex]finally derivating the period with respect to the mass we do:[tex]\frac{dT}{dm} = 2\pi * \frac{d(\sqrt{\frac{m}{k} })}{dm} = 2\pi * \frac{1}{\sqrt{k}}  * \frac{d(\sqrt{m})}{dm}\\ = 2\pi * \frac{1}{2} *\frac{1}{\sqrt{k}}  * \frac{1}{\sqrt{m}} = \frac{\pi }{ \sqrt{m*k}}[/tex]The sign of the last expression tells you that if we decrease the mass or the spring constant the exchange rate of the period with respect to the mass is going to increase