{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics 325 HW 5 Solutions - Physics 325 Fall 2010...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 325 Fall 2010: Solutions to Homework Assignment #5 1) a)Determining x ( t ) and y ( t ) Method 1: The only forces acting on the mass are from the springs. Using Newtons law F x = - kx - kx = m ¨ x m ¨ x = - 2 kx Similarly m ¨ y = - 2 ky . The general solutions for these two differential equations are x ( t ) = A x cos( ωt ) + B x sin( ωt ) y ( t ) = A y cos( ωt ) + B y sin( ωt ) where ω = 2 k m Plug in the initial conditions: x (0) = A x = x 0 ˙ x (0) = ωB x = V x to give A x = x 0 and B x = v 0 x ω and A y = y 0 and B y = v 0 y ω Then the solutions are x ( t ) = x 0 cos( ωt ) + v 0 x ω sin( ωt ) y ( t ) = y 0 cos( ωt ) + v 0 y ω sin( ωt ) Method 2: Since the oscillations in x direction and y direction are decoupled with each other, we only discuss the movement in x direction and similar method applies to the oscillation in y direction. Assume x ( t ) = A cos( ωt - δ ), where ω = 2 k/m , (it is easy to verify that x ( t ) satisfies the equation of motion.) Using the initial condition, x (0) = A cos δ = x 0 ˙ x (0) = A 2 k/m sin δ = v 0
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Solving the equations, we have A = x 2 0 + m 2 k v 2 0 δ = arctan( m 2 k v 0 x 0 ) Method 3: We can verify that the equation of motion has a complex solution z ( t ) = Ae i ( ωt - δ ) where ω = 2 k/m . Let x ( t ) = Re ( z ( t )), use the initial condition that x 0 = x (0) = Re ( z (0)) = A cos δ v 0 = ˙ x (0) = Re ( ˙ z (0)) = Re ( iωAe i ( ωt - δ ) ) = A 2 k/m sin δ So we have A = x 2 0 + m 2 k v 2 0 δ = arctan( m 2 k v 0 x 0 ) and x ( t ) = Rez ( t ) = A cos( ωt - δ ) .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern