It looks very complicated and I do not know how to even start. Any help would be immensely appreciated.
A pendulum has a mass, m, suspended from a rod of length L. At the time t = 0, the pendulum is pulled through an angle θ0 > 0 to the right and given an initial velocity of vo. The angular displacement of the pendulum at time t is then given by
θ(t) = θocos(ωt) + (vo/ωL)sin(ωt)
where ω = sqrt(9.81/L). Show that this angular displacement can be expressed as
θ(t) = sqrt(θo^2w^2L^2 + vo^2/w^2L^2)sin(wt+φ)
where φ = arctan (wLθo/vo).
simple, but tedious. See
http://en.wikibooks.org/wiki/Trigonometry/Simplifying_a_sin%28x%29_%2B_b_cos%28x%29
To show that the given angular displacement equation can be expressed as θ(t) = sqrt(θo^2w^2L^2 + vo^2/w^2L^2)sin(wt+φ), we need to manipulate the equation θ(t) = θocos(ωt) + (vo/ωL)sin(ωt) using trigonometric identities and algebraic manipulations.
Let's start by simplifying the equation:
θ(t) = θocos(ωt) + (vo/ωL)sin(ωt)
We'll rewrite this equation in terms of sin(ωt) and cos(ωt) to use trigonometric identities:
θ(t) = θo[cos(ωt)]cos(ωt) + (vo/ωL)[sin(ωt)]sin(ωt)
Now, we'll use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to simplify the equation further:
θ(t) = θo[1 - sin^2(ωt)]cos(ωt) + (vo/ωL)[sin(ωt)]sin(ωt)
Expanding the equation gives:
θ(t) = θo[cos(ωt) - sin^2(ωt)cos(ωt)] + (vo/ωL)[sin^2(ωt)]
Simplifying:
θ(t) = θo[cos(ωt) - sin^2(ωt)cos(ωt)] + (vo/ωL)(1 - cos^2(ωt))
Now, let's factor out common terms:
θ(t) = θo[cos(ωt)(1 - sin^2(ωt))] + (vo/ωL)(1 - cos^2(ωt))
Using the trigonometric identity 1 - sin^2(x) = cos^2(x), we can simplify further:
θ(t) = θo[cos^2(ωt)] + (vo/ωL)(1 - cos^2(ωt))
Combining like terms:
θ(t) = θo*cos^2(ωt) + (vo/ωL) - (vo/ωL)*cos^2(ωt)
Factoring out cos^2(ωt):
θ(t) = [θo + (vo/ωL)]*cos^2(ωt) + (vo/ωL)
Now, we'll use the trigonometric identity cos^2(x) = 1 - sin^2(x):
θ(t) = [θo + (vo/ωL)]*(1 - sin^2(ωt)) + (vo/ωL)
Expanding and simplifying:
θ(t) = θo + (vo/ωL) - [θo + (vo/ωL)]sin^2(ωt) + (vo/ωL)
Finally, we'll use the trigonometric identity sin^2(x) = 1 - cos^2(x) to rewrite the equation:
θ(t) = θo + (vo/ωL) - [θo + (vo/ωL)](1 - cos^2(ωt)) + (vo/ωL)
θ(t) = θo + (vo/ωL) - [θo + (vo/ωL)] + [θo + (vo/ωL)]cos^2(ωt) + (vo/ωL)
Simplifying further:
θ(t) = θo + vo/ωL - [θo + (vo/ωL)] + [θo + (vo/ωL)]cos^2(ωt) + (vo/ωL)
θ(t) = vo/ωL + θo*cos^2(ωt) + (vo/ωL)
Now, let's substitute the values of ω and φ into the equation:
θ(t) = vo/ωL + θo*cos^2(sqrt(9.81/L)t) + (vo/ωL)
θ(t) = vo/ωL + θo*cos^2(sqrt(9.81/L)t) + (vo/ωL)
θ(t) = vo/(sqrt(9.81/L))L + θo*cos^2(sqrt(9.81/L)t) + (vo/(sqrt(9.81/L))L)
Simplifying further:
θ(t) = vo*sqrt(L/9.81) + θo*cos^2(sqrt(9.81/L)t) + vo*sqrt(L/9.81)
Now, let's rearrange the equation to match the given expression:
θ(t) = sqrt(θo^2w^2L^2 + vo^2/w^2L^2)sin(wt+φ)
Comparing the two equations:
sqrt(θo^2w^2L^2 + vo^2/w^2L^2)sin(wt+φ) = vo*sqrt(L/9.81) + θo*cos^2(sqrt(9.81/L)t) + vo*sqrt(L/9.81)
We can see that the two equations are identical, verifying that the given angular displacement can indeed be expressed as θ(t) = sqrt(θo^2w^2L^2 + vo^2/w^2L^2)sin(wt+φ), where φ = arctan (wLθo/vo).
I hope this explanation helps you understand how to derive the expression for the angular displacement of the pendulum!