a-block-with-a-mass-of-10-mathrmkg-connected-to-a-spring-oscillates-back-and-forth-with-an-amplitude

Question

A block with a mass of 10 $\mathrm{kg}$ connected to a spring oscillates back and forth with an amplitude of 2 $\mathrm{m}$ . What is the approximate period of the block if it has a speed of 4 $\mathrm{m} / \mathrm{s}$ when it passes through its equilibrium point?
(A) 1 $\mathrm{s}$
(B) 3 $\mathrm{s}$
(C) 6 $\mathrm{s}$
(D) 12 $\mathrm{s}$

Sachin Rao · Accepted Answer

problem. We have given a blog with a mask. Dan kg is connected to a spring oscillates back and forth with an amplitude two m. What is the Premix? Approximate period of the vlog if it has a speed four metre pass again when it passes through aquarium point. So we know that at the current point is feeders maximum which is equal to a into omega and it is given to us four m per second and amplitude is also given. So we will right to into America is equal to four. So angular frequency will be equal to to radiant for a second. Okay, Now we have to find that time period. We know that time period is equal to two Pi Omega. We have to find the time period rate approximately time period to five. Iomega. So this is equal to bye by is equal to around 3.14 So option B three seconds will be right. Answer. Thank you

Averell Hause · Answer

we can use for part of the conservation of mechanical energy 1/2 m. Visa of one squared plus 1/2 que delta X quantity squared equal in 1/2 and the initial squared and then initially there isn&#x27;t any. Uh, there there isn&#x27;t any elastic stored energy, so we can then say that 1/2 okay, a squared would be equaling 2 1/2 and the initial squared pass Cancel out and we find that then a is equal in the square root or rather is equaling the initial velocity multiplied by the square root of the mass divided by okay to spring constant. So the amplitude is then equaling two 0.40 meters per second multiplied by 1.0 kilograms. This will be divided by 16 Newtons per meter. And so this is giving us 2.10 meters. Four part B. We can find the velocity at a point where X is Equalling half of the amplitude. And so we have been that 1/2 MVS of two squared, plus 1/2 times k multiplied by a over to quantity squared equals 1/2. I am the initial squared plus zero jewels and so we can say that. Then 1/2 m visa two squared equals 1/2 and the initial squared minus 1/4 multiplied by 1/2. Okay, a squared. And so we can say that this would be Equalling two again. 1/2 and the initial squared minus 1/4 multiplied by one half m the initial square. And we find that then 1/2 m. These of two squared equals 3/4. I multiplied by 1/2, Um, the initial squared. And so we can say that then be some two equals initial velocity multiplied by the square root of 3/4. It&#x27;s a visa to is equaling. Then 0.40 meters per second, multiplied by the square root of 3/4. And this is giving us 0.346 meters per second. So this would be our velocity at half the amplitude. That is the end of the

Salamat Ali · Answer

Antique energy office. Spring mechanic Energy is equal to the potential energy office free so we can write a tactic. Energies one or two and we square from here. We will find amplitude so I can write potential energy off the spring is one or two k a square. So is the amplitude off this spring? So the amplitude came. You found by solving for a year canceling art one over two on both sides. Then the amplitude we get for a spring is in terms. Offer Mars and Velocity and Kay&#x27;s Every Square derided by K been constant. Maas we have here is a tree. We&#x27;re lost days of four. So four square derided by spring constant is a 10 squealed taking swirled off. This number gives us the answer is 2.2 meters, so our options sees the right answer.

Eric Mockensturm · Answer

So I told a 12 kilogram block is supported by a spring shown if the block is moved vertically upward from its equilibrium position and released determined the period of frequency, the resulting motion, the maximum velocity and acceleration of the block if the amplitude of its motion is 50 millimeters. Okay, so we have our spring constant, and our mass and we confined immediately are natural frequency. Just being careful about this killer Newton&#x27;s here and we get 20.41 radiance per second, which then gives us 3.25 hertz and a period of 0.308 seconds, which is simply one over this now, the the, um we have simple harmonic motion again, and so I&#x27;ve again shifted time. So that teak was when t zero x zero. Um, And so in this case, we&#x27;re actually told the see that the, uh the maximum the amplitude of its motion is 50 millimeters, so a is gonna be 50 millimeters. So here&#x27;s our velocity and acceleration. So if a has 50 millimeters or is your 0.5 meters, then our maximum velocity is a times omega en or 1.2 meters per second and a maximum acceleration is a times omega and squared, or 20.8 meters per second squared.

Averell Hause · Answer

So here we have simple harmonic motion and this would be modelled by exposition of the function of time equaling the amplitude times co sign of the angular frequency multiplied by time, plus the face constant. The velocity in the X direction as a function of time would be Equalling to the negative amplitude multiplied by the angular frequency multiplied by sine of the angular frequency multiplied by time plus again the face constant. So we need to then determine the angular frequency, the face constant and the amplitude. So we can first start off by saying the period equaling two pi multiplied by the mass over spring constant K. This would be willing to two pi multiplied by 1.0 kilograms. This would be divided by 20 Newtons per meter and this is giving us 1.405 seconds. So the angular frequency would then be Equalling to two pi divided by t the time period two pi divided by 1.405 seconds. And this is giving us then 4.472 radiance for a second. We know that initially we can say at T equaling zero seconds the initial position equals the amplitude multiplied by co sign uh, find not. And we know that the initial velocity and the extraction would be going too negative. A Omega sign. If I not dividing these two equations, we can say that then tangent of fine not would be equaling being negative initial velocity in the X direction divided by the angular frequency multiplied by the initial exposition. This is equaling two, then negative negative 1.0 meters per second. This would be divided by 4.472 radiance per second multiplied my point 20 meters and we find that then this is equaling 1.1181 and find not would then simply be arc 10 of 1.1181 giving us 0.841 radiance. So we can say that then, from the initial conditions, the amplitude would be equaling square root of the initial position squared plus the initial velocity in the X direction divided by the angular frequency quantity squared. This is equaling than the square root of 0.20 meters. It&#x27;s less negative. 1.0 meters per second, divided by 4.472 radiance per second and this would be squared. Both terms would be square. And we take the square root of the sum of the squares and we find that the amplitude is equaling two point 300 meters. So using this equation all the way up here, here we can graph plugging in the amplitude, the angular frequency in the face constant. This final equation is here, and this would be the grass. That is the end of the solution. Thank you for watching.

Averell Hause · Answer

So here, for part athe damped, angular velocity is going to be equal to the square root of the spring. Constant divided by the mass, uh, minus B squared over for em. And so this is going to be equal to the square root of 41 Newtons per meter, divided by the mass of 0.835 kilograms. And then this will be minus 0.6662 uh, Newton seconds per meter squared. And then this will be divided by four times the mass of 0.835 kilograms, and we find that the damped angular velocity will be equal to 6.996 radiance per second. Now, for part B, they want to find the fractional change. So the fractional change would be equal to the amplitude times e to the negative gamma t AA minus a the amplitude divided by a the amplitude. And so this is gonna be equal to e to the negative gamma capital T period minus one. Um, and then we can say that this is gonna be equal to e x p e x p just is the exponential function. So whatever is in the parentheses is the power with a base of E, so this would be negative b over to em times t and then close parentheses minus one. We can solve and say that this is gonna be the exponential function of negative Ah, 0.6662 and then this will be divided by two times point 835 kilograms minus one and this is giving us negative 0.300 So I would be your answer for part B the fractional change. And then for part C, they want us to find the, um thea position off of a position function as a position function with respect to time. And uh however, the initial position is 0.120 meters. So the general equation for the damp oscillator will be equal to a times e x p of negative gamma times the period brother times t rather times sign of omega prime times t. And so, for the first part, we have 0.120 meters equals a thanks e x p r. The exponential function of negative 0.6662 times one second, divided by two times the mass of 20.835 A sign of 6.996 radiance. And so this is basically except T equals one second equals 10.120 meters. That&#x27;s how we&#x27;re setting this equation off. And solving for a A is simply gonna be equal 2.27 three meters. Ah, you would simply have Thio plug this into your T 80 45 or 89 and soul for a ah, you can use a cell function in order to isolate a so a is gonna be equal 2.273 meters. Gamma is going to be equal to be over to, um and this will be equal to a point 662 divided by two times 20.835 And this is giving us 0.396 per second. And so constructing the position versus time. Rather, position function position as a function of Time X would be equal to X would be equal to 0.273 meters, uh, times the exponent exponential function of negative 0.396 t and then sign of seven t. So this would be the full equation. That is the end of the solution. Thank you for watching

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A block with a mass of 4 $\mathrm{kg}$ is attache…

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