Physics Solution Manual for 1100 and 2101

Thus the speed of the wave is v 2lf1 combining this

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Unformatted text preview: string vibrates at the lower frequency. The waves on the longer string have the same speed as those on the shorter string. The speed v of a transverse wave on a string is given by v = F / ( m / L ) (Equation 16.2), where F is the tension in the string and m/L is the mass per unit length (or linear density). Since F and m/L are the same for both strings, the speed of the waves is the same. 0.57 cm SOLUTION The beat frequency is the frequency of the shorter string minus the frequency of the longer string; fshorter − flonger . We are given that fshorter = 225 Hz. According to Equation 17.3 with n = 1, we have flonger = v/(2Llonger), where Llonger is the length of the longer string. According to the drawing, we have Llonger = L + 0.0057 m. Thus, L Chapter 17 Problems f longer = v 2Llonger = 927 v 2 ( L + 0.0057 m ) Since the speed v of the waves on the longer string is the same as those on the shorter string, v = 41.8 m/s. The length L of the shorter string can be obtained directly from Equation 17.3: L= v 41.8 m/s = = 0.0929 m 2 f1 2 ( 225 Hz ) Substituting this number back into the expression for flonger yields flonger = 41.8 m/s v = = 212 Hz 2 ( L + 0.0057 m ) 2 ( 0.0929 m + 0.0057 m ) The beat frequency is fshorter − flonger = 225 Hz − 212 Hz = 13 Hz . 39. SSM WWW REASONING The natural frequencies of the cord are, according to Equation 17.3, f n = nv / ( 2 L ) , where n = 1, 2, 3, .... The speed v of the waves on the cord is, according to Equation 16.2, v = F / ( m / L ) , where F is the tension in the cord. Combining these two expressions, we have 2 nv n F fn = = 2L 2L m / L fn 2L F = m/ L n or Applying Newton's second law of motion, ΣF = ma, to the forces that act on the block and are parallel to the incline gives F – Mg sin θ = Ma = 0 or F = Mg sin θ where Mg sin θ is the component of the block's weight that is parallel to the incline. Substituting this value for the tension into the equation above gives 2 fn 2L Mg sin θ = m/ L n This expression can be solved for the angle θ and evaluated at the various harmonics. The answer can be chosen from the resulting choices. SOLUTION Solving this result for sin θ shows that 928 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA (m / L) sin θ = fn 2L 1.20 × 10 –2 kg/m (165 Hz ) 2 ( 0.600 m ) 3.20 = = 2 Mg n n n (15.0 kg ) 9.80 m/s 2 2 2 ( ) Thus, we have 3.20 n2 θ = sin –1 Evaluating this for the harmonics corresponding to the range of n from n = 2 to n = 4 , we have 3.20 θ = sin –1 2 = 53.1° for n = 2 2 3.20 = 20.8° for n = 3 32 θ = sin –1 3.20 = 11.5° for n = 4 42 θ = sin –1 The angles between 15.0° and 90.0° are θ = 20.8 ° and θ = 53.1° . 40. REASONING AND SOLUTION We are given that f j = 12 2 f j –1 . a. The length of the unfretted string is L0 = v/(2f0) and the length of the string when it is pushed against fret 1 is L1 = v/(2f1). The distance between the frets is L0 – L1 = v f v v v 1 – = 1– 0 = 1– 2 f 12 2 2 f 2 f 0 2 f1 0 f1 0 1 = L0 1 – 12 = ( 0.628 m )( 0.0561) = 0.0352 m 2 b. The frequencies corresponding to the sixth and seventh frets are f 6 = f7 = (12 2 ) 7 f 0 . The distance between fret 6 and fret 7 is (12 2 ) 6 f 0 and Chapter 17 Problems L6 – L7 = 929 v v – 2 f6 2 f7 v = 2 (12 2 ) = L0 6 v – f0 1 2 – (12 2 ) 1 (12 2 ) (12 2 ) 6 7 7 v = f0 2 f0 1 – 1 (12 2 ) (12 2 ) 6 7 = ( 0.628 m )( 0.0397 ) = 0.0249 m 41. SSM REASONING AND SOLUTION The distance between one node and an adjacent antinode is λ/4. Thus, we must first determine the wavelength of the standing wave. A tube open at only one end can develop standing waves only at the odd harmonic frequencies. Thus, for a tube of length L producing sound at the third harmonic (n = 3), L = 3(λ/4). Therefore, the wavelength of the standing wave is λ= 4L= 3 4 3 1 = b.5 m g 2.0 m and the distance between one node and the adjacent antinode is λ/4 = 0.50 m . 42. REASONING Equation 17.5 (with n = 1) gives the fundamental frequency as f1 = v/(4L), where L is the length of the auditory canal and v is the speed of sound. SOLUTION Using Equation 17.5, we obtain f1 = 343 m/s v = = 3.0 ×103 Hz 4 L 4 ( 0.029 m ) 43. REASONING The frequency of a pipe open at both ends is given by Equation 17.4 as v fn = n , where n is an integer specifying the harmonic number, v is the speed of 2L sound, and L is the length of the pipe. SOLUTION Solving the equation above for L, and recognizing that n = 3 for the third harmonic, we have v L = n 2f n 343 m/s = 3 = 1.96 m 2 ( 262 Hz ) 930 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 44. REASONING The speed v of sound is related to its frequency f and wavelength λ by v = f λ (Equation 16.1). At the end of the tube where the tuning fork is, there is an antinode, because the gas molecules there are free to vibrate. At the plunger, there is a node, because the gas molecules there are not free to vibrate. Since there is an antinode at one end of the tube and a node at the other, the smallest value of L occurs when the length of the tube is one quarter of a wavelen...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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