How Do You Derive Equations (3a) and (4a) for Coupled Pendulums?

[8614smith]

Homework Statement

two coupled pendulums are used at positions x,1 and x,2

Newton’s equation for the forces leads to the two equations:

m,1 * (second derivative of x,1 with respect to t) = -k,1x,1 + k(x,2 - x,1)

and m,2 * (second derivative of x,2 with respect to t) = -k,2x,2 - k(x,2 - x,1)

This leads to the two solutions:

x,1(t) = A,1*sin(ω,1*t + α,1) + A,2*sin(ω,2*t + α,2) (equation 3)

and

x,2(t) = A,1*sin(ω,1*t + α,1) - A,2*sin(ω,2*t + α,2) (equation 4)

where

A,1 = A,2 and α,1 = α,2

rewrite equations (3) and (4) in the very interesting form:

x,1(t) = 2A,1*cos(((ω,1 - ω,2)/2)*t)sin(((ω,1 + ω,2)/2) (equation 3a)

and

x,2(t) = 2A,1*sin(((ω,1 - ω,2)/2)*t)cos(((ω,1 + ω,2)/2) (equation 4a)



Basically i have to derive (3a) and (4a) from equations 3 and 4 using A,1 = A,2 and α,1 = α,2

Homework Equations

above

The Attempt at a Solution

all I've managed to do is expand out the brackets and that's where i get stuck, is there anyone that can help get me in the right direction as i have no idea where to go from here or how it changes from sin to a cos,

thanks

 

[tiny-tim]

Welcome to PF!

Hi 8614smith! Welcome to PF!

You need to learn the four equations for sinA ± sinB and cosA ± cosB.

In this case, use sinA + sin B = 2.sin((A+B)/2).cos((A-B)/2)