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A 1.5 kg object oscillates with simple harmonic motion on spring of force constant k 575 N/m_ Its maximum speed is 42.0 cm/s_What the totab mechanical energy?What t...

Question

A 1.5 kg object oscillates with simple harmonic motion on spring of force constant k 575 N/m_ Its maximum speed is 42.0 cm/s_What the totab mechanical energy?What the amplitude of the oscillation? cm

A 1.5 kg object oscillates with simple harmonic motion on spring of force constant k 575 N/m_ Its maximum speed is 42.0 cm/s_ What the totab mechanical energy? What the amplitude of the oscillation? cm



Answers

$\bullet$ An object of mass $1.0 \mathrm{~kg}$ is attached to a spring with spring constant $15 \mathrm{~N} / \mathrm{m}$. If the object has a maximum speed of $0.50 \mathrm{~m} / \mathrm{s}$, what is the amplitude of oscillation?

This problem goes the concept of the simple harmonic motion and for part a, the mechanical energy is securing the maximum potential energy that is half of the spring constant. K into the squad room temperature and subscribe the value. Half of the spring constant is to 15 or 10 per meter. And then they do this 3.5 centimeter or you can write 0.35 m squared. So the mechanical energy of the system is 0.15 tools. Mhm. Now, partly the maximum speed is the equivalent of the amplitude into square root of the spring constant upon the mass. As I stood the value. So the maximum speed equals 0.035 major into square Rudolph The spring constant to 15, 10 Parameter Upon the Mars, and that is 0.5 kg. So the maximum speech equals 0.7, eight m for second. Okay. Mhm. Now the maximum acceleration for part B, the maximum acceleration equals the omega square or into the spring constant. K. Born the mass m Are the maximum acceleration equals zero 035 Meadows into 2 50 Newton Parameter Upon 0.5 kg are the maximum acceleration is uh 18 medals, four seconds

Solving party of this problem here. The speed can be written had B. X. Is equal to Underwood. Okay but I am multiplication is square minus X squared describing both sides. So I can write the expression like this on. For the simplification I can write the value of A. Is equal to Underwood and by Kay multiplication we exist squared plus X squared. So I will just put the value in the next step. We're just putting the value in this expression. The expression can be written it understood .60 Kg by 14. Newton permitted multiplication .95 m/s. Holy Square Plus .22 m square which is equal to 0.295 m as the amplitude. Now solving part B. I can ride. The value of is equal to one x 2K. A square which is equal to one by two. Multiplication, 14 U. Turn parameter multiplication. 0.295 m police square. On solving it further. Finally I can ride evaluate just look at it carefully. 0.609 jewel as total mechanical energy. Now solving part C. The year I can write the value of P. X. Is called to under route. Hereby AM multiplication is square minus X. Is square. To just putting the value here I can I. D express a net under the hood. 14 x 0.60 multiplication 0.295 meter holy square minus 0.11 m holy square Which is equal to 1.32 m pulse again as our answer.

So here we know that four part athe maximum force is gonna be equal to the mass times acceleration some M. This is simply Newton's second law, and we know that the acceleration amplitude is gonna be equal to Omega Square terms. Except m we know that the angular velocity or the angular frequency with vehicle to two pi times of frequency, we know the frequency is simply the reciprocal of the period. So this is equaling one over a point to this is equaling five hertz. So we can say that the angular velocity is simply 10 pie radiance for a second. And so the maximum force would be equal to the massive point 12 kilograms multiplied by 10 pie radiance per second quantity squared, multiplied by, uh, 0.8 five meters. This is giving us 10 Newtons. This would be our answer for part a. We know that four part B the angular frequency is related by is related to the string constant. The angular frequency is equal to the square root of the spring constant divided by the mass. Therefore, the spring constant is equaling the mass times the angular frequency squared. This would be equal to 0.12 kilograms multiplied by 10 pie radiance for a second quantity squared. And we find that the, uh, spring constant is equal in 1.2 times 10 to the second Newtons per meter. This would be our final answer for part B. That is the end of the solution. Thank you.

Solving party of this problem. Here I can write, the value is equal to uh huh. K A T square. On further simplification, I can write the value of K. Is equal to E by a square. So just putting the value in this ex prison, I can write attached to multiplication. Want to fight you. Bye. 1.5 km Holy Square which is equal to 111.11 Newton Per meter as the spring content. Now I will solve part of me. So here I can write the value of P is equal to two pi route and the mm bikey which is required to to pine root and 0.750 x 111.11 Which is equal to zero 5 2 2nd every time period solving part. See here I can die debate. Your Pemex is equal to too bye bye. T multiplication A which is equal to two pi multiplication, 1.5 metre by You're a .52 seconds, which is equal to 18.12 m/s as the maximum is spilling so I can write the maximum acceleration Mx is equal to to buy by the police square, which is omega is square multiplication A, which is equal to four pi square, a bitey square, Which is equal to four multiplication by square. My duplication 1.5 m by euro 0.52 2nd. Foley square, which is equal to 219 m per second squared as the acceleration of exhilarated.


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