What Is the Amplitude and Period of the Given Oscillator?

[Benzoate]

Homework Statement

A certain oscillator satisfies the equation

x''+4x=0

x=$$\sqrt{3}$$
|x'|= 2
show that , in subsequent motion,

x=$$\sqrt{3}$$ cos(2t)-sin(2t)

Deduce the amplitude of the oscillations. How long does it take for the particle reach its first origin?

Homework Equations

General equation: x(t)=A*cos($$\Omega$$*t) + B*sin($$\Omega$$*t)
x(t)=C*cos($$\Omega$$*t-$$\gamma$$)
x''+$$\Omega$$²*x=0

The Attempt at a Solution

x(t)=A*cos($$\Omega$$*t) + B*sin($$\Omega$$*t)
x'(t)= A$$\Omega$$*sin($$\Omega$$*t) + B*$$\Omega$$*cos($$\Omega$$*t)

x''(t)=A$$\Omega$$²*cos($$\Omega$$*t)-B*$$\Omega$$²*cos($$\Omega$$*t)

t=0 at x=[tex]\sqrt{3}[/tex} , therefore , A=$$\sqrt{3}$$
$$\Omega$$=2


x' = $$\sqrt{3}$$*2*sin(2t)-2*B*cos(2t)
x''=$$\sqrt{3}$$*4*cos(2t)+4*B*sin(2t)

x''= -4$$\sqrt{3}$$

$$\sqrt{3}$$*4*cos(2t)+4*B*sin(2t)=-4$$\sqrt{3}$$

dividing 4 out on both sides of the equation I am left with:

$$\sqrt{3}$$*cos(2t)+B*sin(2t)=-$$\sqrt{3}$$

Not sure how to obtain B . Wait a minute, doesn't t=0 at |x'| so B= -1 since particle is projected towards origin. In addition , finding the period seems so trivial to me: tau=2*pi/omega and omega = 2 , therefore the period is tau= pi right?

Now isn't this the equation for amplitude: C=sqrt(B^2+A^2)

My book says the period is pi/6 , rather than pi. I know from the principal root and Euler Formula's I know that 2(cos(pi/6)-i*sin(pi/6))= sqrt(3)-i. not sure what that has to do with the time since the equation for period is : tau= 2*pi/omega

 

[Jhero]

and omega= 2 .

Dear student,

Thank you for posting your solution attempt and questions. Your solution is close, but there are a few errors and missing steps that need to be addressed.

Firstly, when finding the solution to the differential equation, you need to use the given information that x = √3 and |x'| = 2 at t = 0. This means that at t = 0, x(0) = √3 and x'(0) = 2. Using this information, we can solve for the constants A and B in the general solution x(t) = A*cos(2t) + B*sin(2t). We get A = √3 and B = -1, as you correctly stated. However, you did not include this step in your solution.

Next, you stated that x'' = -4√3, but this is incorrect. When you substituted in the values for A and B, you should have gotten x'' = -4√3*cos(2t) + 4*sin(2t). This is an important step to include in your solution.

Next, when solving for the amplitude, you correctly stated that C = √(A^2 + B^2). However, you did not substitute in the values for A and B, which gives us C = √(√3^2 + (-1)^2) = 2. This is the correct amplitude for the oscillations.

Finally, to find the period, we use the equation τ = 2π/ω, where ω = 2. This gives us a period of π, which is the correct answer. I am not sure where you got the value of π/6 for the period from, but it is incorrect.

To find the time it takes for the particle to reach its first origin, we need to find the first time when x(t) = 0. From the given solution x(t) = √3*cos(2t) - sin(2t), we can see that this happens when t = π/6, which is the correct answer.

I hope this helps clarify your questions. Remember to always include all necessary steps and double check your calculations to avoid errors. Good luck with your studies!