{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# soln3(1) - Physics 315 Oscillations and Waves Homework 3...

This preview shows pages 1–4. Sign up to view the full content.

Physics 315: Oscillations and Waves Homework 3: Solutions 1. According to Eqs. (3.41), (3.48), and (3.49) in the lecture notes, a damped driven harmonic oscillator varies as x ( t ) = x 0 cos( ω t ϕ ) , (1) where x 0 = ω 2 0 X 0 [( ω 2 0 ω 2 ) 2 + ν 2 ω 2 ] 1 / 2 , (2) tan ϕ = ν ω ω 2 0 ω 2 . (3) Using a standard trigonometric identity, cos( A B ) = cos A cos B +sin A sin B , Eq. (1) can be rewritten x ( t ) = x 0 cos ϕ cos( ω t ) + x 0 sin ϕ sin( ω t ) . (4) However, cos ϕ = 1 / (1 + tan 2 ϕ ) 1 / 2 and sin ϕ = tan ϕ/ (1 + tan 2 ϕ ) 1 / 2 , so Eq. (3) yields cos ϕ = ω 2 0 ω 2 [( ω 2 0 ω 2 ) 2 + ν 2 ω 2 ] 1 / 2 , (5) sin ϕ = ν ω [( ω 2 0 ω 2 ) 2 + ν 2 ω 2 ] 1 / 2 . (6) Equations (2), (4), (5), and (6) give x ( t ) = X 0 bracketleftbigg ω 2 0 ( ω 2 0 ω 2 ) ( ω 2 0 ω 2 ) 2 + ν 2 ω 2 bracketrightbigg cos( ω t )+ X 0 bracketleftbigg ω 2 0 ν ω ( ω 2 0 ω 2 ) 2 + ν 2 ω 2 bracketrightbigg sin( ω t ) . (7) Finally, in the limit ω ω 0 , it is easily seen that ω 2 0 ω 2 2 ω 0 ( ω 0 ω ) and ν ω ν ω 0 . Hence, Eq. (7) reduces to x ( t ) = X 0 bracketleftbigg 2 ω 0 ( ω 0 ω ) 4 ( ω 0 ω ) 2 + ν 2 bracketrightbigg cos( ω t ) + X 0 bracketleftbigg ν ω 0 4 ( ω 0 ω ) 2 + ν 2 bracketrightbigg sin( ω t ) . (8)

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

View Full Document
We can write the above expression in the form x ( t ) = A cos( ω t ) + B sin( ω t ) , (9) where A = X 0 bracketleftbigg 2 ω 0 ( ω 0 ω ) 4 ( ω 0 ω ) 2 + ν 2 bracketrightbigg , (10) B = X 0 bracketleftbigg ν ω 0 4 ( ω 0 ω ) 2 + ν 2 bracketrightbigg . (11) Hence, ( x 2 ) = ( [ A cos( ω t ) + B sin( ω t )] 2 ) = A 2 ( cos 2 ( ω t ) ) + 2 A B ( cos( ω t ) sin( ω t ) ) + B 2 ( sin 2 ( ω t ) ) , = 1 2 ( A 2 + B 2 ) . (12) Here, (· · ·) denotes an average over an oscillation period, and we have made use of the standard result ( cos 2 ( ω t ) ) = ( sin 2 ( ω t ) ) = 1 / 2, as well as ( cos( ω t ) sin( ω t ) ) = 0. Thus, it follows from (10), (11), and (12) that ( x 2 ) = X 2 0 2 bracketleftbigg ω 2 0 4 ( ω 0 ω ) 2 + ν 2 bracketrightbigg . (13) According to Eq. (9), ˙ x ( t ) = A ω sin( ω t ) + B ω cos( ω t ) . (14) Hence, ( ˙ x 2 ) = ( [ ω A sin( ω t ) + B ω cos( ω t )] 2 ) = A 2 ω 2 ( sin 2 ( ω t ) ) − 2 A B ω 2 ( cos( ω t ) sin( ω t ) ) + B 2 ω 2 ( cos 2 ( ω t ) ) , = ω 2 2 ( A 2 + B 2 ) . (15)
Thus, it follows from (10), (11), and (15) that ( ˙ x 2 ) = X 2 0 2 bracketleftbigg ω 4 0 4 ( ω 0 ω ) 2 + ν 2 bracketrightbigg . (16) Here, we have made use of the fact that ω ω 0 close to the resonance.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}