P2214HW7 - PHYS2214, Spring 2009 Solutions to Assignment 7...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
PHYS2214, Spring 2009 Solutions to Assignment 7 March 23, 2009 1 Energy and power in sound waves a) As was derived in class, the average intensity of a sound wave is related to the molecular displacement amplitude s m and the pressure amplitude p m by I av = 1 2 Bkωs 2 m (1) = B 2 v ω 2 s 2 m (2) = v 2 B p 2 m (3) Therefore for a fixed average intensity s m = r 2 vI av B 1 ω (4) 1 ω (5) and p m = r 2 BI av v (6) ω 0 (7) b) We have 2 ω = 1 KHz sound waves, one propogating in water ( B w = 2 . 2 × 10 9 Pa and v w = 1480 m/s) and one in air ( B a = 1 . 4 × 10 5 Pa and v a = 343 m/s), having equal I av . Using eqn.[4], we see that the ratio of the molecular diplacements is s w m s a m = r v w B a v a B w (8) = 0 . 0166 (9) Likewise from eqn.(6), we have p w m p a m = r v a B w v w B a (10) = 60 . 35 (11) 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
We see that p w m p a m = s a m s w m p w m s w m = p a m s a m . Since also, I av = 1 2 ωp m s m (12) and ω ’s are identical, this I w av = I a av (13) c) The percieved loudness ( β ) of a soundwave is related to it’s intensity by β = 10 log 10 I I 0 (14) where I 0 = 10 - 12 W / m 2 , is the threshold of hearing. β is expressed in decibels (dB). 1) We have a 1 W point source of waves, i.e the total energy crossing a sphere of radius r about this point, per sec is 1 J. The intensity, which is power/unit area is hence I ( r ) = 1 4 πr 2 W / m 2 (15) The difference in loudness between a point a distance r from the source and one that is twice as far away is δβ = β ( r ) - β (2 r ) (16) = 10(log 10 I ( r ) I 0 - log 10 I (2 r ) I 0 ) (17) = 10 log 10 I ( r ) I (2 r ) (18) = 10 log 10 4 (19) = 6 . 02 (20) The initial ditance from the source does not affect the change in loudness. 2) The intensity as a function of distance is as in eqn.[13]. The sound level as a function of distance is β ( r ) = 10 log 10 I ( r ) I 0 (21) = 10 log 10 ± 1 4 πr 2 10 - 12 ² (22) = 10 log 10 ± 10 12 4 πr 2 ² (23) (24) We invert the above expression to get the distance at which the intensity is a particular value. r ( β ) = s 1 4 π 10 12 10 β ( r ) 10 (25) r (120dB) = 0 . 28 m (26) r (65dB) = 158 . 63 m (27) r (20dB) = 28209 m (28) (29) 2
Background image of page 2
120 dB is the threshold of pain. This says that you are at the threshold of pain if you are 28 cm away from a speaker that emits 1 W
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/01/2009 for the course PHYS 2214 taught by Professor Giambattista,a during the Spring '07 term at Cornell University (Engineering School).

Page1 / 7

P2214HW7 - PHYS2214, Spring 2009 Solutions to Assignment 7...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online