ps02soln - PHYSICS 218 SOLUTION TO HW 2 Created September 3...

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PHYSICS 218 SOLUTION TO HW 2 Created: September 3, 2004 11:38pm Last updated: September 12, 2004 7:46pm 1. The mass per unit length is given by (refer to assignment 1) λ 0 = 1 4 π D 2 = 6 . 283 × 10 - 3 kg/m , where and D are respectively the density and diameter of the string. Assuming that the string is oscillating in its fundamental mode with A = 2 mm and frequency 261 . 6 Hz, the total vibrational energy of the string is therefore E = 1 4 λ 0 ω 2 n A 2 = 1 4 × 0 . 604 × (6 . 283 × 10 - 3 ) × (2 π × 261 . 6) 2 × (2 × 10 - 3 ) 2 = 1 . 025 × 10 - 2 J . 2. Our tasks here is to verify the following orthogonality relations for sin( ω n t ) and cos( ω n t ), T 1 0 sin( m ω 1 t ) sin( n ω 1 t ) d t = T 1 2 δ mn , (1) T 1 0 cos( m ω 1 t ) cos( n ω 1 t ) d t = T 1 2 δ mn , (2) T 1 0 sin( m ω 1 t ) cos( n ω 1 t ) d t = 0 , (3) where T 1 is the period of the fundamental, T 1 = 2 π ω 1 . Let’s begin with the case m = n . The usual way is to express sin 2 ( n ω 1 t ) and cos( n ω 1 t ) as 1 2 (1 - cos(2 1 t )) and 1 2 (1 + cos(2 1 t )) respectively and evaluate the integral using the preliminary knowledge we have about integrals of sin(2 n ω 1 t ) and cos(2 n ω 1 t ). An alternative way of doing it is to observe the fact that the integral of sin 2 ( n ω 1 t ) within any interval of π is actually identical to the integral of cos 2 ( n ω 1 t ) within the same interval π , namely, τ + T 1 τ sin 2 ( n ω 1 t ) d t = τ + T 1 τ cos 2 ( n ω 1 t ) d t , (4) where τ is any constant. It is easy to see this by just looking and comparing the graphs of sin 2 ϑ and cos 2 ϑ , as shown in figure (1). Equation (4) simply states that for any interval α ϑ α + π , the area under both curves are the same. A more rigorous way of proving the aforementioned statement is by explicit calculations. For the sake of not muddling up our current preoccupation, let’s defer it and we shall come back to this point later. Equation (4) can be extended to any interval with integer multiples of π to obtain 2 0 sin 2 ϑ d ϑ = 2 0 cos 2 ϑ d ϑ . 1
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PHYSICS 218 SOLUTION TO HW 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . π 2 π 3 π 2 2 π ϑ 0 1 0 1 sin 2 ϑ cos 2 ϑ Figure 1: The graph of sin 2 ϑ and cos 2 ϑ over the interval 0 ϑ 2 π . Making a change of variable ϑ = n ω 1 t , and using the identity sin 2 ϑ + cos 2 ϑ = 1, we find 2 π/ω 1 0 sin 2 ( n ω 1 t ) d t = 1 n ω 1 2 0 sin 2 ϑ d ϑ = 1 2 n ω 1 2 0 ( sin 2 ϑ + cos 2 ϑ ) d ϑ = 1 2 n ω 1 2 πn 0 d ϑ = π ω 1 = T 1 2 .
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