Study Guide 2 - Acoustics

(a) This is a Displacement Antinode, because the molecules are free to move there. Since the minimum number of nodes possible is 1, then the standing wave pattern must be: 



NOTE: This is NOT a displacement node because the molecules are free to move at these positions. At a displacement node there is no motion. In this case there must be displacement antinodes at each end.

(b) The tube is closed at the right end and open at the left. What is at each end?

• (1) node - node: No: Molecules are not free to move at the right end but are free to move at the left end. There must be a displacement node at the right and a displacement antinode at the left.
• (2) node - antinode: No: Molecules are not free to move at the right end but are free to move at the left end. There must be a displacement node at the right and a displacement antinode at the left.
• (3) antinode - node: Yes. The molecules are free to move at the left end (displacement antinode) and cannot move at the right end (displacement node). The standing wave pattern must be:
  
  
   
• (4) antinode - antinode: No: Molecules are not free to move at the right end but are free to move at the left end. There must be a displacement node at the right and a displacement antinode at the left.

(c) This is a Displacement Node, because the molecules are not free to move there so there must be a displacement node at each end. The minimum number of antinodes between 2 nodes is 1 so the standing wave pattern must be:



NOTE: This is NOT a displacement antinode. The molecules are not free to move. At a displacement antinode the molecules must be free to move. There must be a displacement node at each end.

The next highest frequency will have one more node in each case so the standing wave patterns will be:

Let us first look at the cases in problem 1. Since the length of the tube is 1 m:


 

(a) The distance between two adjacent resonances (nodes) is 1/2 wavelength so
\(\mathrm {\lambda =20 \;cm = 0.20 \;m}\)

(b) \(\mathrm {f = v/\lambda = 340/0.20 = 1.70\times 10^3 \;Hz}\)

(c) No: It would just shift the pattern by 1/4 \(\lambda\) making the right end a node instead of an antinode.

Beats occur at the differences of the two frequencies involved. However we don't know if the upper C is at a higher or a lower frequency than 512 Hz. So the possibilities are:

512 + 5 = 517 Hz
512 - 5 = 507 Hz

(a) \(\mathrm {\log\; 2.00 = 0.301}\)

(b) \(\mathrm {\log \;10 = 1}\)

(c) \(\mathrm {\log\; 1.13 = 0.053}\)

(d)  \(\mathrm {\log \;(1.40\times 10^6) = 6.146}\)
You should notice that it is also
\(\mathrm {6 + \log\; 1.40 = 6 + 0.146 = 6.146}\)

(e) \(\mathrm {\log \;(0.000670) = -3.174}\)
You should notice that it is also
\(\mathrm {-4 + \log \; 6.70 = -4 + 0.826 = -3.174}\)

(f) \(\mathrm {\log \;(9.40\times10^{-6}) = -5.027}\)
You should also notice that it is
\(\mathrm {-6 + \log \;9.40 = -6 + 0.973 = -5.027}\)

(g) \(\mathrm {\log \;(94.0\times10^{-7}) = \log\; (9.40\times 10^{-6})}\)
so the answer is the same as for Question 6. (f) above

Note: The number of significant figures in a logarithm is only the number of places after the decimal point since the integer part only gives the power of 10. All the numbers in the question (except (b)) are given to 3 figures and so the answers are given to 3 figures.

(a) \(\mathrm {antilog \;(0.724) = 5.30}\)
(b) \(\mathrm {antilog\;(1.724) = 53.0}\)
(c) \(\mathrm {antilog\; (-0.689) = 0.205}\)
(d) \(\mathrm {antilog\; (-6.89) = 1.3\times 10^{-7}}\)
(e) \(\mathrm {antilog\; (-0.0100) = 0.977}\)

Note: The only significant figures in a logarithm are those after the decimal piont. The integer part just gives the power of 10. In the following there are 3 significant figures in (a), (b), (c) and (e); (d) has 2 sig. fig. This must be preserved in the answer.

Powers like this can be solved directly on most calculators by using the \(\mathrm {y^x}\) key.
If you don't have a \(\mathrm {y^x}\) key, they can be solved by taking logarithms:

(a) \(\mathrm {\log \;(4.00^{0.360}) = 0.360 \log \;4.00 = 0.360\times 0.602 = 0.2167}\)
\(\mathrm {antilog \;(0.2167) = 1.65}\)

(b) \(\mathrm {(7.10)^{\log\;3.20} = 7.10^{0.505}}\)
\(\mathrm {\log\; (7.10^{0.505}) = 0.505 \log\; 7.10 = 0.505\times 0.851 = 0.4299}\)
\(\mathrm {antilog\; (0.4299) = 2.69}\)

(c) \(\mathrm {(\log \;2.40)^{\log 0.930} = 0.380^{-0.0315}}\)
\(\mathrm {\log \;(0.380^{-0.0315}) = -0.0315\; \log \;0.380 = -0.0315(-0.4202) = 0.01324}\)
\(\mathrm {antilog (0.01324) = 1.03}\)

These equations can only be solved by taking logarithms of both sides of the equation.

(a) \(\mathrm {42.0^{(z - 3)} = 7.00}\)

\(\mathrm {\log [42.0^{(z-3)} ] = log \;7.00}\)
\(\mathrm {(z - 3) 1.623 = 0.845}\)
\(\mathrm {z - 3 = 0.521}\)
\(\mathrm {z = 3.52 (3 \;sig. \;fig)}\)

(b) \(\mathrm {4.0(8.0)^z = 7.0(3.0)^{(z - 1)}}\)

\(\mathrm {\log \;4.0 + z\; \log \;8.0 = \log \;7.0 + (z - 1) \log \;3.0}\)
\(\mathrm {0.602 + 0.903\; z = 0.845 + 0.477 (z - 1)}\)
\(\mathrm {0.602 + 0.903 \;z = 0.845 + 0.477 \;z - 0.477}\)
\(\mathrm{0.426 \;z = -0.234}\)
\(\mathrm {z = -0.55 \;(2\; sig. \;fig)}\)
 

It will help to simplify the numerator and denominator separately and then take the ratio.

Simplify numerator
\(\mathrm {\log (ab^2) + log (1/a^2b)}\)
\(\mathrm {= \log (ab^2) + log (a^{-2}b^{-1})}\)
\(\mathrm {= \log \;a + 2\; \log\; b -2 \; \log \;a - \log \;b}\)
\(\mathrm {= -\log \;a + \log \;b}\)
\(\mathrm {= \log \;(b/a)}\)

Simplify denominator
\(\mathrm {4 \; \log [(1/a^3)(b)(1/a^{-2})]}\)
\(\mathrm {= 4\; log\; (a^{-1} b) = 4 \;log\; (b/a)}\)

\(\mathrm {Ratio = [\log\; (b/a)]/[4\; \log\;(b/a)] = 1/4}\)

e - None of the above!

(a) \(\mathrm {6.84\times10^{-6} W/m^2}\)
\(\mathrm {IL = 10\;\log \;(I/I_0)}\)
\(\mathrm {where\; I_0 = 10^{-12} W/m^2}\)
\(\mathrm {IL = 10 \; \log (6.84\times 10^{-6}/10^{-12})}\)
\(\mathrm {= 10 \; \log \;6.84\times 10^6}\)
\(\mathrm {= 10 (6.835) = 68.4 \;db}\)

(b) \(\mathrm {45.6\times 10^{-14} W/m^2}\)
\(\mathrm {IL = 10 \; \log (I/I_0)}\)
\(\mathrm {where\; I_0 = 10^{-12} \;W/m^2}\)
\(\mathrm {IL = 10\; \log \;(45.6\times 10^{-14}/10^{-12})}\)
\(\mathrm{= 10 \; \log \; 45.6\times 10^{-2}}\)
\(\mathrm {= 10 \;(-0.341) = -3.41\; db}\)

(a) \(\mathrm {-18.0 \;db}\)
\(\mathrm {IL = 10\; \log\; (I/I_0)}\)
\(\mathrm {where\; I_0 = 10^{-12} \;W/m^2}\)
\(\mathrm {-18.0 = 10\; \log\; (I/10^{-12})}\)
\(\mathrm {\log \;(I/10^{-12}) = -1.8}\)
\(\mathrm {I/10^{-12} = 0.0158 \;I = 1.58\times 10^{-14} \;W/m^2}\)

(b) \(\mathrm {118 \;db}\)
\(\mathrm {IL = 10\; \log\; (I/I_0)}\)
\(\mathrm {where\; I_0 = 10^{-12}\; W/m^2}\)
\(\mathrm {118 = 10\; \log \;(I/10^{-12})}\)
\(\mathrm {\log\; (I/10^{-12}) = 11.8}\)
\(\mathrm {I/10^{-12} = 6.3\times 10^{11}}\)
\(\mathrm {I = 0.63 \;W/m^2}\)

\(\mathrm {IL_1 = 10 \; \log \; (I_1/I_0)}\)
\(\mathrm {IL_2 = 10\; \log\; (I_2/I_0)}\)
\(\mathrm {IL_1 - IL_2 = 1\;0\ log \; [(I_1/I_0) - (I_2/I_0)]}\)
\(\mathrm {= 10\; \log \;(I_1/I_2)\; (Remember\; the \;difference\; of\; logs\; is \;the\; log \;of\; the \;quotient)}\)
\(\mathrm {105 = 10\; \log\; (I_1/I_2)}\)
\(\mathrm {\log \;(I_1/I_2) = 10.5}\)
\(\mathrm {(I_1/I_2) = 3.2\times 10^{10}}\)

Their intensity.

\(\mathrm {80 = 10\; \log \;(I/10^{-12})}\)
\(\mathrm {log\; (I/10^{-12}) = 8}\)
\(\mathrm {I/10^{-12} = 1\times 10^8}\)
\(\mathrm {I = 1\times 10^{-4} \;W/m^2}\)

For 3 speakers:
\(\mathrm {3I = 3\times 1 \times 10^{-4} = 3\times 10^{-4} \;W/m^2}\)
\(\mathrm {IL = 10\; \log \; (3\times 10^{-4} /10^{-12})}\)
\(\mathrm {= 10\; log\; 3\times 10^8 =10 \times 8.48 = 85 \;db}\)

C 

At about 3500 Hz.

(a) 10 watts

(b) \(\mathrm {Area\; of\; sphere = 4\pi \; r^2 = 4\pi(1)^2 = 12.6 \;m^2}\)
\(\mathrm {P/A = 10W/12.6\; m^2 = 0.796 \;W/m^2}\)
\(\mathrm {Therefore \;the\; power \;on\; 1\; m^2 \;is\; 0.796 \;W.}\)

(c) The intensity of the sound is just the power per unit area, so the answer is the answer to part (b)

(d) \(\mathrm {Energy = Power \times time = 10 \times 2 = 20\; j}\)

(a) \(\mathrm {As\; in\; Q1(a)\; P = 10 \;W}\)

(b) There are 2 ways to solve this:

1. \(\mathrm {I =P/A = 10/[4\pi (10)^2] = 7.96\times10^{-3} W/m^2}\)

2. Remembering from Q18 (b) that the intensity at 1 m was \(\mathrm {0.796\; W/m^2}\) and using the inverse square law:
\(\mathrm {I =P/A = 0.796/10^2 = 7.96\times 10^{-3}\; W/m^2}\)

(c) From part (b) \(\mathrm {I = 7.96\times 10^{-3} \;W/m^2}\)
\(\mathrm {Area = 0.10 \times 0.10 = 0.010 \;m^2}\)
\(\mathrm{Power = IA = 7.96\times 10^{-3} (W/m^2) \times 0.010\; m^2}\)
\(\mathrm {= 7.96\times 10^{-5} \; W}\)

(d) From part (c) , \(\mathrm {Power\; on\; square = 7.96\times 10^{-5}\; W}\)
\(\mathrm {Energy = Power \times time = 7.96\times 10^{-5} \times 3.00 = 2.39\times 10^{-4}\; j}\)

(a) Auditory Canal
(b) Tympanic Membrane
(c) Ossicles
(d) Basilar Membrane
(e) Helicotrema