Physics I Chapter L Solutions

Chapter L - Problems Problem L.1 Consider a pulse that in SI units has the shape 8 u = f HxL = x2 +4 . Write this as a function uHx, tL that describes this pulse moving in the positive x direction with a speed of 3 m ê s. Solution to L.1 A pulse of shape u = f HxL moving in the positive x-direction with speed v takes the form: u = f Hx - v tL. Using this form of f HxL with v = 3 m ê s gives: f HxL = 8 x2 +4 ï u= 8 Hx - 3 tL2 +4 . Problem L.2 What are the speed and direction of a pulse on a string that (in SI units) has the form: yHx, tL = 0.04 e -J x + 0.03 t 2 0.06 N . Solution to L.2 A wave of the form: yHx, tL = f Hx ° v tL represents a pulse of shape y = f HxL moving in the ± x direction with speed v. Here we have 2 yHx, tL = 0.04 e-Hxê0.06L with v = 0.03 m ê s moving in the negative x direction. Problem L.3 A sinusoidal pulse on a string has the mathematical form yHx, tL = H0.80 mL sinA 2p 10 t = 0.6 s how much has the pulse shifted. On the same graph plot y vs. x at t = 0.6 s. Solution to L.3 After t = 0.6 s the graph has shifted by v t = 4 µ 0.6 = 2.4 m. y HmL 0.75 0.5 0.25 5 -0.25 -0.5 -0.75 10 15 20 x HmL Hx - 4 tLE. Plot the y vs. x graph at t = 0 s. By the time 2 Chapter K - Problems Problem L.4 A string with a linear density of m = 4 µ 10-3 kg ê m is given a tension of 360 N. What is the speed of waves on this string? Solution to L.4 The speed of waves on a stretched string is v = Here, T = 360 N and m = 4 µ 10 -3 T ê m where T is the tension in the string and m is the linear density (mass/length) of the string. kg ê m giving v = 300 m s . Problem L.5 The elastic limit for steel is Smax = 2.7 µ 109 N ê m2 ; this is the maximum force per area that steel under tension can withstand. Smax is the largest value that T ê A, the tension per area, can have without a wire breaking. If the density of steel is 7860 kg ê m3 then what is the largest speed a wave can travel down a steel wire? Solution to L.5 The linear density m (mass/length) is related to the volume density r (mass/volume) by l = r A, where A is the cross-sectional area of the wire. This is easy to show: The volume of a wire of length L and area A is L A. Mass = r µ Volume = r L A and l = Mass ê L = r A. The maximum stress Stressmax gives the maximum tension: Tmax = Smax µ A, which then gives the miximum wave speed. vmax = Tmax m Smax A = = rA Smax r = 2.7µ109 = 586 7.86µ103 m s Problem L.6 A 30 m long copper wire with a 1.2 mm diameter is stretched to a tension of 200 N. How long does it take for a pulse to travel the length of the wire? The density of copper is r = 8.92 µ 103 kg ê m3 . Solution to L.6 The cross-sectional area of the wire is A = p r2 = p µ 0.00062 = 1.1310 µ 10-6 m2 . We saw in the previous problem that the linear density m of a wire with density r and cross-sectional area A is m = r A. The speed of a wave on a wire with tension T, cross-section A and density r is v= T m T = rA = 200 8.92µ103 A = 141 m s . To get the time from the distance d and speed simply use d = v t giving: t= d v = 30 v = 0.213 s. Problem L.7 A sinusoidal wave on a string has the form yHxL = H15 cmL cosAI p 20 cm-1 M x - H16 p s-1 L tE. (a) Plot the motion of the positon x = 0 as a function of time and find its period and frequency. (b) What is the maximum speed of this point (x = 0) on the string? (c) What are the wavelength and speed of this wave? Solution to L.7 Chapter K - Problems 3 (a) At x = 0 we have yHxL = A cosHw tL, where A = 15 cm and w = 16 p s-1 . The period T and frequency f are T= 2p w = 1 8 w = 0.125 s and f = 2p = 1 T = 8 Hz. With this we can plot the function. y HcmL 15 10 5 -5 0.05 0.1 0.15 0.2 t HsL -10 -15 (b) This motion is an example of simple harmonic motion, where the maximum speed is given by vmax = w A = 16 p µ 15 = 240 p = 754 cm s . (c) The wave number k of this wave gives the wavelength. k= p 20 cm-1 ï l = 2p = 40 cm k The speed can be found by using v = f l or directly in term of what is giving by using v= w k 16 p = pê20 m = 320 s . Problem L.8 As a sinusoidal wave passes, a point on a string makes 50 complete vibrations is 20 s. In the same time a crest (maximun) of the wave moves a distance of 4 m. What is the frequency, speed and wavelength of this wave? Solution to L.8 50 vibrations in 20s gives a frequency of f= 50 20 = 2.5 Hz . 4 m in 20 s gives the speed. v= 4 = 0.2 20 m s We can now find the wavelength. f l=v ï l= v f = 0.2 2.5 = 0.08 m Problem L.9 A 15 m length of rope has a mass of 0.6 kg and is given a tension of 500 N. What power is required to put a wave with an amplitude of 20 cm and a frequency of 3 Hz? Solution to L.9 The linear density is m = M ê L = 0.6 ê 15 = 0.04 kg ê m. The speed of the wave is v= T m = 500 0.04 = 111.80 m ê s 4 Chapter K - Problems T v= m 500 = 0.04 = 111.80 m ê s The angular frequency follows from the frequency. w = 2 p f = 2 p 3 = 18.850 s-1 Using A = 0.20 m we can now find the power transmitted down a string. P = 1 2 m w2 A2 v = 1 2 0.04 µ 1.88502 H0.20L2 111.80 = 31.8 W. Problem L.10 A wave of the form yHx, tL = H0.12 mL sin@H0.8 m-1 L x +H40 s-1 L tD travels down a string with a linear density of 8 g ê m. (a) What is the speed of the wave and in what direction is it moving? (b) What are the wavelength and frequency of this wave? (c) What is the tension in the string? (d) What is the power transmitted by this wave? Solution to L.10 Since the general form is yHx, tL = A sinHk x ° w t -fL , we can conclude that: A = 0.12 m , k = 0.8 m-1 and w = 40 s-1 . We are also given that m = 0.008 kg m . (a) The positive sign before w implies it is moving in the negative x direction. The speed is v= w k = 40 0.8 = 50 m s (b) The wavelength and frequency come from the wave number and angular freauency. l= 2p k = 2p 0.8 = 7.85 m and f = w 2p = 40 2p = 6.37 Hz (c) The speed and linear density give the tension. v= T m ï T = v2 m = 502 µ 0.008 = 20 N (d) The power transmitted is P = 1 2 m w2 A2 v = 1 2 0.008 µ 402 H0.12L2 50 = 4.61 W.