1.) A sailor strikes the side of his ship just below the waterline. He hears the

Question

1.) A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor directly below 2.2 s later. How deep is the ocean at this point? Assume the speed of sound in seawater is 1560 m/s (Table 12-1) and does not vary significantly with depth.

_____ m

2.) What is the intensity of sound at 104 dB?

____ W/m2

Compare it to that of a whisper at 20 dB.

____ W/m2

3.) If you were to build a pipe organ with open-tube pipes spanning the range of human hearing (20 Hz to 20 kHz), what would be the range of the lengths of pipes required?

shortest pipe

______ m

longest pipe

______ m

4.) The predominant frequency of a certain fire engine's siren is 1400 Hz when at rest. What frequency do you detect if you move with a speed of 24.0 m/s in the following directions.

(a) toward the fire engine

_____ Hz

(b) away from the it

_____ Hz

5.) A bat flies toward a wall at a speed of 4.5 m/s. As it flies, the bat emits an ultrasonic sound wave with frequency 31.5 kHz. What frequency does the bat hear in the reflected wave?

_____ kHz

Explanation / Answer

1. D = S x T

D = 1560(Speed of Sound) x 1.1(It takes 2.2 to go to the ocean floor and back again to the ears of the sailor. We need to use the time for sound to go to the ocean floor once)

D = 1716m

2.

3.L=nv/2f where n=1, 2, 3, etc.

Long
L=343/40= 8.58m

Short
S-343/40000= .00858m

4.assuming 0 degree temp.
toward fire siren
Fa = Fn ( Va +Vi / Va)

= 1400 Hz ( 331 m/s + 24 m/s / 331 m/s)
= 1400 Hz (355 m/s / 331 m/s)
= 1400 (1.0725)
= 1501.51 Hz

away from fire siren
Fa = Fn (Va - Vi / Va)

=1400 Hz (331 m/s - (-24 m/s) / 331 m/s)
=1400 Hz (355 m/s / 331 m/s)
= 1400 (1.0725)
= 1501.51 Hz

5.The bat will hear a higher frequency than it emitted,due to the Doppler effect
The formula for this:
f_new = [(c + V_r)/(c + V_s)]*f_o

- f_o is the original frequency
- c is the speed of sound
- V_r is the velocity of the receiver relative to the air (positive if moving TOWARDS the source)
- V_s is the velocity of the source relative to the air (positive if moving AWAY from the receiver)

a) Stage 1: The source is the bat, the receiver is the wall:
f_ o = 33 (kHz)
V_r = 0
V_s = 10 (m/s)
c = 343 (m/s)

f_new = [(c + V_r)/(c + V_s)] * f_o
= [(343 + 0)/(343 - 10)] * 33
= [343/(333)] * 33
= 1.03 * 33
= 33.99 (kHz)

b) Stage 2: the source is the wall, the receiver is the bat:
f_ o' = 33.99 (kHz)
V_r = 10 (m/s)
V_s = 0
c = 343 (m/s)

f_new' = [(c + 10)/(c + 0] * f_o'
= [(343 + 10)/343] * 33.99
= 34.98 (kHz)

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