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SolSec3_3 - 3 37 Problems and Solutions Section...

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3- 37 Problems and Solutions Section 3.3 (problems 3.26-3.32) 3.26 Derive equations (3.24). (3.25) and (3.26) and hence verify the equations for the Fourier coefficient given by equations (3.21), (3.22) and (3.23). Solution: For n ! m , integration yields: 0 T ! sin n " T t sin m " T tdt = sin n # m ( ) " T t " T 2 n # m ( ) # sin n + m ( ) " T t " T 2 n + m ( ) \$ % & & ( ) ) 0 T = sin n # m ( ) 2 * T + , - . / 0 T \$ % & ( ) 2 n # m ( ) " T # sin n + m ( ) 2 * T + , - . / 0 T \$ % & ( ) 2 n + m ( ) " T = sin n # m ( ) 2 * ( ) \$ % ( 2 n # m ( ) " T # sin n + m ( ) 2 * ( ) \$ % ( 2 n + m ( ) " T = 0 Since m and n are integers, the sine terms are 0, so this is equal to 0. Equation (3.24), for m = n : 0 T ! sin 2 n " T tdt = 1 2 t # 1 4 n " T sin 2 n " T t ( ) \$ % & ( ) 0 T = T 2 # T 8 n * sin 2 * 2 * T + , - . / 0 T \$ % & ( ) = T 2 # T 8 n * sin 4 n * \$ % ( = T 2 Since n is an integer, the sine term is 0, so this is equal to T /2. So, 0 T ! sin n " T t sin m " T tdt = 0 m # n T / 2 m = n \$ % & Equation (3.25), for m ! n

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3- 38 0 T ! cos n " T t cos m " T tdt = sin n # m ( ) " T t 2 n # m ( ) " T # sin n + m ( ) " T t 2 \$ + m ( ) " T % & ( ) * * 0 T = sin n # m ( ) 2 + T , - . / 0 1 T % & ( ) * 2 n # m ( ) " T # sin n + m ( ) 2 + T , - . / 0 1 T % & ( ) * 2 n + m ( ) " T = sin n # m ( ) 2 + ( ) % & ( ) 2 n # m ( ) " T # sin n + m ( ) 2 + ( ) % & ( ) 2 n + m ( ) " T = 0 Since m and n are integers, the sine terms are 0, so this is equal to 0. Equation (3.25), for m = n becomes: cos 2 0 T ! n " T tdt = 1 2 t + 1 4 n " T sin 2 n " T t ( ) # \$ % & ( 0 T = T 2 + T 8 n ) sin 2 n 2 ) T * + , - . / T # \$ % & ( = T 2 + T 8 n ) sin 4 n ) # \$ & = T 2 Since n is an integer, the sine term is 0, so this is equal to T /2. So, 0 T ! cos n " T t cos m " T tdt = 0 m # n T / 2 m = n \$ % & Equation (3.26), for m ! n : 0 T ! cos n " T t sin m " T tdt = cos n # m ( ) " T t 2 " T n # m ( ) # cos n + m ( ) " T t 2 " T n + m ( ) \$ % & & ( ) ) 0 T = cos n # m ( ) 2 * T + , - . / 0 T \$ % & ( ) 2 n # m ( ) " T # cos n + m ( ) 2 * T + , - . / 0 T \$ % & ( ) 2 n + m ( ) " T # 1 2 m # n ( ) " T + 1 2 m + n ( ) " T = cos n # m ( ) 2 * ( ) \$ % ( 2 n # m ( ) " T # cos n + m ( ) 2 * ( ) \$ % ( 2 n + m ( ) " T # 1 2 m # n ( ) " T + 1 2 m + n ( ) " T = 0 Since n is an integer, the cosine term is 1, so this is equal to 0.
3- 39 So, 0 T ! cos n " T t sin m " T tdt = 0 Equation (3.26) for n = m becomes: 0 T ! cos n " T t sin n " T tdt = 1 2 n " T sin 2 n " T t # \$ % & ( 0 T = T 4 n ) sin 2 2 ) n = 0 Thus 0 T ! cos n " T t sin n " T tdt = 0

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3- 40 3.27 Calculate b n from Example 3.3.1 and show that b n = 0, n = 1,2,…, for the triangular force of Figure 3.12. Also verify the expression a n by completing the integration indicated. ( Hint : Change the variable of integration from t
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