Problem Set 9 Solutions

Problem Set 9 Solutions - Problem Set IX Solutions Fall...

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Unformatted text preview: Problem Set IX Solutions Fall 2006 Physics 200a 1. Write the equation for a wave moving along + x with amplitude .4 m , speed 6m/s and frequency 17 Hz . If these are waves on a string with mass per unit length = . 02kg/m , what is the u , the energy per unit length? What is the power being fed into the vibrating string? Equation (16-2) gives us the general equation for a moving wave: ( x , t ) = A cos(2 x / + 2 ft ) We will take the sign since we want a wave moving along the + x direction. But, we were given the wave velocity and frequency, not wavelength, so we must use Equation (16-1): v = f Solving this for and then substituting into the chosen form of Equation (16-2) gives: ( x , t ) = A cos(2 xf / v- 2 ft ) ( x , t ) = (0.4 m ) cos[2 (17 Hz ) x /(6 m/ s) - 2 (17 Hz ) t ] ( x , t ) = (0.4 m ) cos[(17 /3 m-1 ) x (34 Hz ) t ] From the lecture notes, we know that the energy per unit length for a sinusoidal wave on a string is: u = A 2 2 This is given in angular frequency, but we only have the frequency. Fortunately, Equation (15-7) reminds us: f = / 2 or 2 f = Which gives the final result that: u = 2 2 A 2 f 2 u = 2 2 (. 02kg/m )(0.4 m ) 2 (17 Hz ) 2 18.3 J / m The power that must be fed into the string varies with what part of wave being produced, but the average power is given by Equation (16-8): < P > = 2 A 2 v This is again given in angular frequency when we only have the frequency. Again making use of Equation (15-7) lets us say that the average power is: < P > = (2 f ) 2 A 2 v = 2 2 f 2 A 2 v < P > = 2 2 (0.02 kg/m )(17 Hz ) 2 (0.4 m ) 2 (6 m / s ) 109.5 W 2. The speed of sound in water and air is 1450 m/s and 330 m/s respectively. Sound from an explosion on the surface of a lake first reaches me when my head is underwater and 5 s later when my head is above the water. How far away was the explosion? From equation (2-1) we know that t = D / v where t is time spent moving, D is the total distance moved and v is the velocity when moving. From the problem we can construct the following two equations: (1) t air = D / v air (2) t water = D / v water But we also know that t air = t water + t where t is 5 s, the difference in travel times. So, we can rewrite our equations as: (1*) t water + t = D / v air or t water = ( D / v air ) - t (2*) t water = D / v water If we then combine the second form of (1*) and (2*) we get: ( D / v air ) t = D / v water D v water- t v air v water = D v air D ( v water- v air ) = t v air v water D = ( t v air v water ) / ( v water- v air ) Evaluating with the given numeric values gives: D = (5 s )(330 m/s )(1450 m/s ) / (1450 m/s - 330 m/s ) = 2136.16 m 2140 m 3. A block of mass M sits on a frictionless inclined plane of angle = /4 as in Figure (1). It is connected by a wire of linear mass density...
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This note was uploaded on 07/13/2008 for the course PHYS 200 taught by Professor Ramamurtishankar during the Fall '08 term at Yale.

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Problem Set 9 Solutions - Problem Set IX Solutions Fall...

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