2.  In this section we will show the gruesome details in the derivation of the speed of sound in liquids:  In a taut string, potential energy is associated with the periodic stretching of tensioned string elements. Topic 11.1 Extended B – The speed of sound v = BB The speed of Sound in a Liquid. B is bulk modulus  In a fluid, potential energy is associated with the periodic expansions and contractions of small volume elements  V of the fluid.  The bulk modulus is defined as ratio of the change in pressure  p to the corresponding fractional change in volume  V/V: B = - pV/VpV/V The Bulk Modulus of a Liquid  The ratio  V/V has no units, so the units if B are those of pressure: n/m 2. FYI: The harder a liquid is to compress, the smaller the fractional change in volume. How does this affect the bulk modulus? FYI: The harder a liquid is to compress, the larger the bulk modulus. How does this affect the speed of sound through that liquid? FYI: The minus sign in the formula for bulk modulus ensures that B is always positive. Can you see how?

3.  Consider a simplified pulse of fluid moving down the tube as shown: Topic 11.1 Extended B – The speed of sound  As the compression zone moves through the fluid at the wave velocity v, it runs into non-compressed fluid. compression zone  We now analyze (using Newton's 2nd law) a small element of that non-compressed fluid, shown immediately to the right of the compression zone: v xx A xx A mass element  The mass of the non-compressed fluid element is given by mass = density·volume m =  V m =  ·  x·A m =  A  x

4. Topic 11.1 Extended B – The speed of sound  All of the fluid within the mass element is accelerated to the wave speed v in the time it takes the compression zone to move through the distance  x. But compression zone v xx A mass element m =  A  x v = xtxt so that  x = v  t which we may substitute: m =  A  x m =  Av  t

5. Topic 11.1 Extended B – The speed of sound  We know that the average acceleration of the mass element is compression zone v xx A mass element m =  Av  t a = vtvt so that Newton's 2nd law,  F = m a, becomes  F = m a  F = (  Av  t) vtvt  F = (  Av)(  v)  F = (  Av 2 )(  v/v) Why?

6. Topic 11.1 Extended B – The speed of sound  Recall that pressure = force/area, so that compression zone v xx mass element F = pA  F = (  Av 2 )(  v/v)  The fluid element therefore has two forces acting on it in the direction of its acceleration, shown above. (p +  p)A pApA  Thus the sum of the forces acting on the fluid element is given by  F = (p +  p)A - pA  F = A  p and we can substitute:  F = (  Av 2 )(  v/v) A  p = (  Av 2 )(  v/v)  F = pA +  pA - pA  p = (  v 2 )(  v/v)

7. Topic 11.1 Extended B – The speed of sound  Just as we could express the mass in terms of the wave velocity, we can express the volume V in terms of the wave velocity v: compression zone v xx mass element  p = (  v 2 )(  v/v) V = A  x V = Av  t  V = A  v  t  If the volume of the mass element is V = Av  t, then the change in volume of the mass element is  Since the change in volume of the mass element is negative, but its change in velocity is positive, we put a negative sign in the formula to get  V = -A  v  t

8. Topic 11.1 Extended B – The speed of sound compression zone v xx mass element  p = (  v 2 )(  v/v)  Now we can look at the fractional change in volume of the fluid element, and write it in terms of the wave velocity:  V = -A  v  t V = Av  t VVVV = -A  v  t Av  t = - vvvv so that  v/v = -  V/V

9. Topic 11.1 Extended B – The speed of sound compression zone v xx mass element  p = (  v 2 )(  v/v)  Substitution yields  v/v = -  V/V  p = (  v 2 )(  v/v)  p = (  v 2 )(-  V/V)  v 2 = - pV/VpV/V  v 2 = B v = BB